818 lines
24 KiB
Markdown
818 lines
24 KiB
Markdown
<p align="center">
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<a href="https://programmercarl.com/other/xunlianying.html" target="_blank">
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<img src="../pics/训练营.png" width="1000"/>
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</a>
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<p align="center"><strong><a href="https://mp.weixin.qq.com/s/tqCxrMEU-ajQumL1i8im9A">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们收益！</strong></p>
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# 704. 二分查找
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[力扣题目链接](https://leetcode.cn/problems/binary-search/)
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给定一个 n 个元素有序的（升序）整型数组 nums 和一个目标值 target ，写一个函数搜索 nums 中的 target，如果目标值存在返回下标，否则返回 -1。
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示例 1:
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```
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输入: nums = [-1,0,3,5,9,12], target = 9
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输出: 4
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解释: 9 出现在 nums 中并且下标为 4
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```
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示例 2:
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```
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输入: nums = [-1,0,3,5,9,12], target = 2
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输出: -1
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解释: 2 不存在 nums 中因此返回 -1
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```
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提示：
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* 你可以假设 nums 中的所有元素是不重复的。
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* n 将在 [1, 10000]之间。
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* nums 的每个元素都将在 [-9999, 9999]之间。
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## 算法公开课
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**[《代码随想录》算法视频公开课](https://programmercarl.com/other/gongkaike.html)：[手把手带你撕出正确的二分法](https://www.bilibili.com/video/BV1fA4y1o715)，相信结合视频再看本篇题解，更有助于大家对本题的理解**。
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## 思路
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**这道题目的前提是数组为有序数组**，同时题目还强调**数组中无重复元素**，因为一旦有重复元素，使用二分查找法返回的元素下标可能不是唯一的，这些都是使用二分法的前提条件，当大家看到题目描述满足如上条件的时候，可要想一想是不是可以用二分法了。
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二分查找涉及的很多的边界条件，逻辑比较简单，但就是写不好。例如到底是 `while(left < right)` 还是 `while(left <= right)`，到底是`right = middle`呢，还是要`right = middle - 1`呢？
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大家写二分法经常写乱，主要是因为**对区间的定义没有想清楚，区间的定义就是不变量**。要在二分查找的过程中，保持不变量，就是在while寻找中每一次边界的处理都要坚持根据区间的定义来操作，这就是**循环不变量**规则。
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写二分法，区间的定义一般为两种，左闭右闭即[left, right]，或者左闭右开即[left, right)。
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下面我用这两种区间的定义分别讲解两种不同的二分写法。
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### 二分法第一种写法
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第一种写法，我们定义 target 是在一个在左闭右闭的区间里，**也就是[left, right] （这个很重要非常重要）**。
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区间的定义这就决定了二分法的代码应该如何写，**因为定义target在[left, right]区间，所以有如下两点：**
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* while (left <= right) 要使用 <= ，因为left == right是有意义的，所以使用 <=
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* if (nums[middle] > target) right 要赋值为 middle - 1，因为当前这个nums[middle]一定不是target，那么接下来要查找的左区间结束下标位置就是 middle - 1
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例如在数组：1,2,3,4,7,9,10中查找元素2，如图所示：
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代码如下：（详细注释）
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```CPP
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// 版本一
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class Solution {
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public:
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int search(vector<int>& nums, int target) {
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int left = 0;
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int right = nums.size() - 1; // 定义target在左闭右闭的区间里，[left, right]
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while (left <= right) { // 当left==right，区间[left, right]依然有效，所以用 <=
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int middle = left + ((right - left) / 2);// 防止溢出 等同于(left + right)/2
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if (nums[middle] > target) {
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right = middle - 1; // target 在左区间，所以[left, middle - 1]
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} else if (nums[middle] < target) {
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left = middle + 1; // target 在右区间，所以[middle + 1, right]
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} else { // nums[middle] == target
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return middle; // 数组中找到目标值，直接返回下标
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}
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}
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// 未找到目标值
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return -1;
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}
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};
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```
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* 时间复杂度：O(log n)
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* 空间复杂度：O(1)
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### 二分法第二种写法
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如果说定义 target 是在一个在左闭右开的区间里，也就是[left, right) ，那么二分法的边界处理方式则截然不同。
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有如下两点：
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* while (left < right)，这里使用 < ,因为left == right在区间[left, right)是没有意义的
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* if (nums[middle] > target) right 更新为 middle，因为当前nums[middle]不等于target，去左区间继续寻找，而寻找区间是左闭右开区间，所以right更新为middle，即：下一个查询区间不会去比较nums[middle]
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在数组：1,2,3,4,7,9,10中查找元素2，如图所示：（**注意和方法一的区别**）
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代码如下：（详细注释）
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```CPP
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// 版本二
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class Solution {
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public:
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int search(vector<int>& nums, int target) {
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int left = 0;
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int right = nums.size(); // 定义target在左闭右开的区间里，即：[left, right)
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while (left < right) { // 因为left == right的时候，在[left, right)是无效的空间，所以使用 <
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int middle = left + ((right - left) >> 1);
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if (nums[middle] > target) {
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right = middle; // target 在左区间，在[left, middle)中
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} else if (nums[middle] < target) {
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left = middle + 1; // target 在右区间，在[middle + 1, right)中
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} else { // nums[middle] == target
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return middle; // 数组中找到目标值，直接返回下标
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}
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}
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// 未找到目标值
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return -1;
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}
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};
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```
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* 时间复杂度：O(log n)
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* 空间复杂度：O(1)
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## 总结
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二分法是非常重要的基础算法，为什么很多同学对于二分法都是**一看就会，一写就废**？
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其实主要就是对区间的定义没有理解清楚，在循环中没有始终坚持根据查找区间的定义来做边界处理。
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区间的定义就是不变量，那么在循环中坚持根据查找区间的定义来做边界处理，就是循环不变量规则。
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本篇根据两种常见的区间定义，给出了两种二分法的写法，每一个边界为什么这么处理，都根据区间的定义做了详细介绍。
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相信看完本篇应该对二分法有更深刻的理解了。
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## 相关题目推荐
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* [35.搜索插入位置](https://programmercarl.com/0035.搜索插入位置.html)
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* [34.在排序数组中查找元素的第一个和最后一个位置](https://programmercarl.com/0034.%E5%9C%A8%E6%8E%92%E5%BA%8F%E6%95%B0%E7%BB%84%E4%B8%AD%E6%9F%A5%E6%89%BE%E5%85%83%E7%B4%A0%E7%9A%84%E7%AC%AC%E4%B8%80%E4%B8%AA%E5%92%8C%E6%9C%80%E5%90%8E%E4%B8%80%E4%B8%AA%E4%BD%8D%E7%BD%AE.html)
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* [69.x 的平方根](https://leetcode.cn/problems/sqrtx/)
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* [367.有效的完全平方数](https://leetcode.cn/problems/valid-perfect-square/)
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## 其他语言版本
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### **Java：**
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（版本一）左闭右闭区间
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```java
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class Solution {
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public int search(int[] nums, int target) {
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// 避免当 target 小于nums[0] nums[nums.length - 1]时多次循环运算
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if (target < nums[0] || target > nums[nums.length - 1]) {
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return -1;
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}
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int left = 0, right = nums.length - 1;
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while (left <= right) {
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int mid = left + ((right - left) >> 1);
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if (nums[mid] == target)
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return mid;
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else if (nums[mid] < target)
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left = mid + 1;
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else if (nums[mid] > target)
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right = mid - 1;
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}
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return -1;
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}
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}
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```
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（版本二）左闭右开区间
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```java
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class Solution {
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public int search(int[] nums, int target) {
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int left = 0, right = nums.length;
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while (left < right) {
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int mid = left + ((right - left) >> 1);
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if (nums[mid] == target)
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return mid;
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else if (nums[mid] < target)
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left = mid + 1;
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else if (nums[mid] > target)
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right = mid;
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}
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return -1;
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}
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}
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```
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### **Python：**
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（版本一）左闭右闭区间
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```python
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class Solution:
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def search(self, nums: List[int], target: int) -> int:
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left, right = 0, len(nums) - 1 # 定义target在左闭右闭的区间里，[left, right]
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while left <= right:
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middle = left + (right - left) // 2
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if nums[middle] > target:
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right = middle - 1 # target在左区间，所以[left, middle - 1]
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elif nums[middle] < target:
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left = middle + 1 # target在右区间，所以[middle + 1, right]
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else:
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return middle # 数组中找到目标值，直接返回下标
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return -1 # 未找到目标值
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```
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（版本二）左闭右开区间
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```python
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class Solution:
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def search(self, nums: List[int], target: int) -> int:
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left, right = 0, len(nums) # 定义target在左闭右开的区间里，即：[left, right)
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while left < right: # 因为left == right的时候，在[left, right)是无效的空间，所以使用 <
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middle = left + (right - left) // 2
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if nums[middle] > target:
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right = middle # target 在左区间，在[left, middle)中
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elif nums[middle] < target:
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left = middle + 1 # target 在右区间，在[middle + 1, right)中
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else:
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return middle # 数组中找到目标值，直接返回下标
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return -1 # 未找到目标值
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```
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### **Go：**
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（版本一）左闭右闭区间
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```go
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func search(nums []int, target int) int {
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high := len(nums)-1
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low := 0
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for low <= high {
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mid := low + (high-low)/2
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if nums[mid] == target {
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return mid
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} else if nums[mid] > target {
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high = mid-1
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} else {
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low = mid+1
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}
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}
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return -1
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}
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```
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（版本二）左闭右开区间
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```go
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func search(nums []int, target int) int {
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high := len(nums)
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low := 0
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for low < high {
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mid := low + (high-low)/2
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if nums[mid] == target {
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return mid
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} else if nums[mid] > target {
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high = mid
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} else {
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low = mid+1
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}
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}
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return -1
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}
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```
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### **JavaScript:**
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（版本一）左闭右闭区间 [left, right]
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```js
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/**
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* @param {number[]} nums
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* @param {number} target
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* @return {number}
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*/
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var search = function(nums, target) {
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// right是数组最后一个数的下标，num[right]在查找范围内，是左闭右闭区间
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let mid, left = 0, right = nums.length - 1;
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// 当left=right时，由于nums[right]在查找范围内，所以要包括此情况
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while (left <= right) {
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// 位运算 + 防止大数溢出
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mid = left + ((right - left) >> 1);
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// 如果中间数大于目标值，要把中间数排除查找范围，所以右边界更新为mid-1；如果右边界更新为mid，那中间数还在下次查找范围内
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if (nums[mid] > target) {
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right = mid - 1; // 去左面闭区间寻找
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} else if (nums[mid] < target) {
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left = mid + 1; // 去右面闭区间寻找
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} else {
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return mid;
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}
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}
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return -1;
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};
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```
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（版本二）左闭右开区间 [left, right)
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```js
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/**
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* @param {number[]} nums
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* @param {number} target
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* @return {number}
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*/
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var search = function(nums, target) {
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// right是数组最后一个数的下标+1，nums[right]不在查找范围内，是左闭右开区间
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let mid, left = 0, right = nums.length;
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// 当left=right时，由于nums[right]不在查找范围，所以不必包括此情况
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while (left < right) {
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// 位运算 + 防止大数溢出
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mid = left + ((right - left) >> 1);
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// 如果中间值大于目标值，中间值不应在下次查找的范围内，但中间值的前一个值应在；
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// 由于right本来就不在查找范围内，所以将右边界更新为中间值，如果更新右边界为mid-1则将中间值的前一个值也踢出了下次寻找范围
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if (nums[mid] > target) {
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right = mid; // 去左区间寻找
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} else if (nums[mid] < target) {
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left = mid + 1; // 去右区间寻找
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} else {
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return mid;
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}
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}
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return -1;
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};
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```
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### **TypeScript**
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（版本一）左闭右闭区间
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```typescript
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function search(nums: number[], target: number): number {
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let mid: number, left: number = 0, right: number = nums.length - 1;
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while (left <= right) {
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// 位运算 + 防止大数溢出
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mid = left + ((right - left) >> 1);
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if (nums[mid] > target) {
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right = mid - 1;
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} else if (nums[mid] < target) {
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left = mid + 1;
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} else {
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return mid;
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}
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}
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return -1;
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};
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```
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（版本二）左闭右开区间
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```typescript
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function search(nums: number[], target: number): number {
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let mid: number, left: number = 0, right: number = nums.length;
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while (left < right) {
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// 位运算 + 防止大数溢出
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mid = left +((right - left) >> 1);
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if (nums[mid] > target) {
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right = mid;
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} else if (nums[mid] < target) {
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left = mid + 1;
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} else {
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return mid;
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}
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}
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return -1;
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};
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```
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### **Ruby:**
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```ruby
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# （版本一）左闭右闭区间
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def search(nums, target)
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left, right = 0, nums.length - 1
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while left <= right # 由于定义target在一个在左闭右闭的区间里，因此极限情况下存在left==right
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middle = (left + right) / 2
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if nums[middle] > target
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right = middle - 1
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elsif nums[middle] < target
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left = middle + 1
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else
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return middle # return兼具返回与跳出循环的作用
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end
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end
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-1
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end
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# （版本二）左闭右开区间
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def search(nums, target)
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left, right = 0, nums.length
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while left < right # 由于定义target在一个在左闭右开的区间里，因此极限情况下right=left+1
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middle = (left + right) / 2
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if nums[middle] > target
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right = middle
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elsif nums[middle] < target
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left = middle + 1
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else
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return middle
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end
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end
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-1
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end
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```
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### **Swift:**
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```swift
|
||
// （版本一）左闭右闭区间
|
||
func search(nums: [Int], target: Int) -> Int {
|
||
// 1. 先定义区间。这里的区间是[left, right]
|
||
var left = 0
|
||
var right = nums.count - 1
|
||
|
||
while left <= right {// 因为taeget是在[left, right]中，包括两个边界值，所以这里的left == right是有意义的
|
||
// 2. 计算区间中间的下标（如果left、right都比较大的情况下，left + right就有可能会溢出）
|
||
// let middle = (left + right) / 2
|
||
// 防溢出：
|
||
let middle = left + (right - left) / 2
|
||
|
||
// 3. 判断
|
||
if target < nums[middle] {
|
||
// 当目标在区间左侧，就需要更新右边的边界值，新区间为[left, middle - 1]
|
||
right = middle - 1
|
||
} else if target > nums[middle] {
|
||
// 当目标在区间右侧，就需要更新左边的边界值，新区间为[middle + 1, right]
|
||
left = middle + 1
|
||
} else {
|
||
// 当目标就是在中间，则返回中间值的下标
|
||
return middle
|
||
}
|
||
}
|
||
|
||
// 如果找不到目标，则返回-1
|
||
return -1
|
||
}
|
||
|
||
// （版本二）左闭右开区间
|
||
func search(nums: [Int], target: Int) -> Int {
|
||
var left = 0
|
||
var right = nums.count
|
||
|
||
while left < right {
|
||
let middle = left + ((right - left) >> 1)
|
||
|
||
if target < nums[middle] {
|
||
right = middle
|
||
} else if target > nums[middle] {
|
||
left = middle + 1
|
||
} else {
|
||
return middle
|
||
}
|
||
}
|
||
|
||
return -1
|
||
}
|
||
|
||
```
|
||
|
||
### **Rust:**
|
||
|
||
```rust
|
||
# （版本一）左闭右闭区间
|
||
|
||
impl Solution {
|
||
pub fn search(nums: Vec<i32>, target: i32) -> i32 {
|
||
let mut left:usize = 0;
|
||
let mut right:usize = nums.len() - 1;
|
||
while left as i32 <= right as i32{
|
||
let mid = (left + right) / 2;
|
||
if nums[mid] < target {
|
||
left = mid + 1;
|
||
} else if nums[mid] > target {
|
||
right = mid - 1;
|
||
} else {
|
||
return mid as i32;
|
||
}
|
||
}
|
||
-1
|
||
}
|
||
}
|
||
|
||
# （版本二）左闭右开区间
|
||
|
||
impl Solution {
|
||
pub fn search(nums: Vec<i32>, target: i32) -> i32 {
|
||
let mut left:usize = 0;
|
||
let mut right:usize = nums.len();
|
||
while left < right {
|
||
let mid = (left + right) / 2;
|
||
if nums[mid] < target {
|
||
left = mid + 1;
|
||
} else if nums[mid] > target {
|
||
right = mid;
|
||
} else {
|
||
return mid as i32;
|
||
}
|
||
}
|
||
-1
|
||
}
|
||
}
|
||
```
|
||
|
||
### **C:**
|
||
|
||
```c
|
||
// (版本一) 左闭右闭区间 [left, right]
|
||
int search(int* nums, int numsSize, int target){
|
||
int left = 0;
|
||
int right = numsSize-1;
|
||
int middle = 0;
|
||
//若left小于等于right，说明区间中元素不为0
|
||
while(left<=right) {
|
||
//更新查找下标middle的值
|
||
middle = (left+right)/2;
|
||
//此时target可能会在[left,middle-1]区间中
|
||
if(nums[middle] > target) {
|
||
right = middle-1;
|
||
}
|
||
//此时target可能会在[middle+1,right]区间中
|
||
else if(nums[middle] < target) {
|
||
left = middle+1;
|
||
}
|
||
//当前下标元素等于target值时，返回middle
|
||
else if(nums[middle] == target){
|
||
return middle;
|
||
}
|
||
}
|
||
//若未找到target元素，返回-1
|
||
return -1;
|
||
}
|
||
```
|
||
```C
|
||
// (版本二) 左闭右开区间 [left, right)
|
||
int search(int* nums, int numsSize, int target){
|
||
int length = numsSize;
|
||
int left = 0;
|
||
int right = length; //定义target在左闭右开的区间里，即：[left, right)
|
||
int middle = 0;
|
||
while(left < right){ // left == right时，区间[left, right)属于空集，所以用 < 避免该情况
|
||
int middle = left + (right - left) / 2;
|
||
if(nums[middle] < target){
|
||
//target位于(middle , right) 中为保证集合区间的左闭右开性，可等价为[middle + 1,right)
|
||
left = middle + 1;
|
||
}else if(nums[middle] > target){
|
||
//target位于[left, middle)中
|
||
right = middle ;
|
||
}else{ // nums[middle] == target ，找到目标值target
|
||
return middle;
|
||
}
|
||
}
|
||
//未找到目标值，返回-1
|
||
return -1;
|
||
}
|
||
```
|
||
|
||
### **PHP:**
|
||
|
||
```php
|
||
// 左闭右闭区间
|
||
class Solution {
|
||
/**
|
||
* @param Integer[] $nums
|
||
* @param Integer $target
|
||
* @return Integer
|
||
*/
|
||
function search($nums, $target) {
|
||
if (count($nums) == 0) {
|
||
return -1;
|
||
}
|
||
$left = 0;
|
||
$right = count($nums) - 1;
|
||
while ($left <= $right) {
|
||
$mid = floor(($left + $right) / 2);
|
||
if ($nums[$mid] == $target) {
|
||
return $mid;
|
||
}
|
||
if ($nums[$mid] > $target) {
|
||
$right = $mid - 1;
|
||
}
|
||
else {
|
||
$left = $mid + 1;
|
||
}
|
||
}
|
||
return -1;
|
||
}
|
||
}
|
||
```
|
||
|
||
### **C#:**
|
||
|
||
```csharp
|
||
//左闭右闭
|
||
public class Solution {
|
||
public int Search(int[] nums, int target) {
|
||
int left = 0;
|
||
int right = nums.Length - 1;
|
||
while(left <= right){
|
||
int mid = (right - left ) / 2 + left;
|
||
if(nums[mid] == target){
|
||
return mid;
|
||
}
|
||
else if(nums[mid] < target){
|
||
left = mid+1;
|
||
}
|
||
else if(nums[mid] > target){
|
||
right = mid-1;
|
||
}
|
||
}
|
||
return -1;
|
||
}
|
||
}
|
||
|
||
//左闭右开
|
||
public class Solution{
|
||
public int Search(int[] nums, int target){
|
||
int left = 0;
|
||
int right = nums.Length;
|
||
while(left < right){
|
||
int mid = (right - left) / 2 + left;
|
||
if(nums[mid] == target){
|
||
return mid;
|
||
}
|
||
else if(nums[mid] < target){
|
||
left = mid + 1;
|
||
}
|
||
else if(nums[mid] > target){
|
||
right = mid;
|
||
}
|
||
}
|
||
return -1;
|
||
}
|
||
}
|
||
```
|
||
|
||
### **Kotlin:**
|
||
|
||
```kotlin
|
||
class Solution {
|
||
fun search(nums: IntArray, target: Int): Int {
|
||
// leftBorder
|
||
var left:Int = 0
|
||
// rightBorder
|
||
var right:Int = nums.size - 1
|
||
// 使用左闭右闭区间
|
||
while (left <= right) {
|
||
var middle:Int = left + (right - left)/2
|
||
// taget 在左边
|
||
if (nums[middle] > target) {
|
||
right = middle - 1
|
||
}
|
||
else {
|
||
// target 在右边
|
||
if (nums[middle] < target) {
|
||
left = middle + 1
|
||
}
|
||
// 找到了，返回
|
||
else return middle
|
||
}
|
||
}
|
||
// 没找到，返回
|
||
return -1
|
||
}
|
||
}
|
||
```
|
||
|
||
### **Kotlin:**
|
||
|
||
```Kotlin
|
||
// （版本一）左闭右开区间
|
||
class Solution {
|
||
fun search(nums: IntArray, target: Int): Int {
|
||
var left = 0
|
||
var right = nums.size // [left,right) 右侧为开区间，right 设置为 nums.size
|
||
while (left < right) {
|
||
val mid = (left + right) / 2
|
||
if (nums[mid] < target) left = mid + 1
|
||
else if (nums[mid] > target) right = mid // 代码的核心，循环中 right 是开区间，这里也应是开区间
|
||
else return mid
|
||
}
|
||
return -1 // 没有找到 target ，返回 -1
|
||
}
|
||
}
|
||
// （版本二）左闭右闭区间
|
||
class Solution {
|
||
fun search(nums: IntArray, target: Int): Int {
|
||
var left = 0
|
||
var right = nums.size - 1 // [left,right] 右侧为闭区间，right 设置为 nums.size - 1
|
||
while (left <= right) {
|
||
val mid = (left + right) / 2
|
||
if (nums[mid] < target) left = mid + 1
|
||
else if (nums[mid] > target) right = mid - 1 // 代码的核心，循环中 right 是闭区间，这里也应是闭区间
|
||
else return mid
|
||
}
|
||
return -1 // 没有找到 target ，返回 -1
|
||
}
|
||
}
|
||
```
|
||
### **Scala:**
|
||
|
||
（版本一）左闭右闭区间
|
||
```scala
|
||
object Solution {
|
||
def search(nums: Array[Int], target: Int): Int = {
|
||
var left = 0
|
||
var right = nums.length - 1
|
||
while (left <= right) {
|
||
var mid = left + ((right - left) / 2)
|
||
if (target == nums(mid)) {
|
||
return mid
|
||
} else if (target < nums(mid)) {
|
||
right = mid - 1
|
||
} else {
|
||
left = mid + 1
|
||
}
|
||
}
|
||
-1
|
||
}
|
||
}
|
||
```
|
||
（版本二）左闭右开区间
|
||
```scala
|
||
object Solution {
|
||
def search(nums: Array[Int], target: Int): Int = {
|
||
var left = 0
|
||
var right = nums.length
|
||
while (left < right) {
|
||
val mid = left + (right - left) / 2
|
||
if (target == nums(mid)) {
|
||
return mid
|
||
} else if (target < nums(mid)) {
|
||
right = mid
|
||
} else {
|
||
left = mid + 1
|
||
}
|
||
}
|
||
-1
|
||
}
|
||
}
|
||
```
|
||
**Dart:**
|
||
|
||
|
||
|
||
```dart
|
||
（版本一）左闭右闭区间
|
||
class Solution {
|
||
int search(List<int> nums, int target) {
|
||
int left = 0;
|
||
int right = nums.length - 1;
|
||
while (left <= right) {
|
||
int middle = ((left + right)/2).truncate();
|
||
switch (nums[middle].compareTo(target)) {
|
||
case 1:
|
||
right = middle - 1;
|
||
continue;
|
||
case -1:
|
||
left = middle + 1;
|
||
continue;
|
||
default:
|
||
return middle;
|
||
}
|
||
}
|
||
return -1;
|
||
}
|
||
}
|
||
|
||
（版本二）左闭右开区间
|
||
class Solution {
|
||
int search(List<int> nums, int target) {
|
||
int left = 0;
|
||
int right = nums.length;
|
||
while (left < right) {
|
||
int middle = left + ((right - left) >> 1);
|
||
switch (nums[middle].compareTo(target)) {
|
||
case 1:
|
||
right = middle;
|
||
continue;
|
||
case -1:
|
||
left = middle + 1;
|
||
continue;
|
||
default:
|
||
return middle;
|
||
}
|
||
}
|
||
return -1;
|
||
}
|
||
}
|
||
```
|
||
|
||
<p align="center">
|
||
<a href="https://programmercarl.com/other/kstar.html" target="_blank">
|
||
<img src="../pics/网站星球宣传海报.jpg" width="1000"/>
|
||
</a>
|
||
|