180 lines
4.8 KiB
Markdown
180 lines
4.8 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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# 283. 移动零：动态规划：一样的套路，再求一次完全平方数
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[力扣题目链接](https://leetcode.cn/problems/move-zeroes/)
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给定一个数组 nums，编写一个函数将所有 0 移动到数组的末尾，同时保持非零元素的相对顺序。
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示例:
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输入: [0,1,0,3,12]
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输出: [1,3,12,0,0]
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说明:
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必须在原数组上操作，不能拷贝额外的数组。
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尽量减少操作次数。
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## 思路
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做这道题目之前，大家可以做一做[27.移除元素](https://programmercarl.com/0027.移除元素.html)
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这道题目，使用暴力的解法，可以两层for循环，模拟数组删除元素（也就是向前覆盖）的过程。
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好了，我们说一说双指针法，大家如果对双指针还不熟悉，可以看我的这篇总结[双指针法：总结篇！](https://programmercarl.com/双指针总结.html)。
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双指针法在数组移除元素中，可以达到O(n)的时间复杂度，在[27.移除元素](https://programmercarl.com/0027.移除元素.html)里已经详细讲解了，那么本题和移除元素其实是一个套路。
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**相当于对整个数组移除元素0，然后slowIndex之后都是移除元素0的冗余元素，把这些元素都赋值为0就可以了**。
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如动画所示：
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C++代码如下：
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```CPP
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class Solution {
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public:
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void moveZeroes(vector<int>& nums) {
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int slowIndex = 0;
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for (int fastIndex = 0; fastIndex < nums.size(); fastIndex++) {
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if (nums[fastIndex] != 0) {
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nums[slowIndex++] = nums[fastIndex];
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}
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}
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// 将slowIndex之后的冗余元素赋值为0
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for (int i = slowIndex; i < nums.size(); i++) {
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nums[i] = 0;
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}
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}
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};
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```
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## 其他语言版本
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### Java：
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```java
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public void moveZeroes(int[] nums) {
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int slow = 0;
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for (int fast = 0; fast < nums.length; fast++) {
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if (nums[fast] != 0) {
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nums[slow++] = nums[fast];
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}
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}
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// 后面的元素全变成 0
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for (int j = slow; j < nums.length; j++) {
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nums[j] = 0;
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}
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}
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```
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### Python：
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```python
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def moveZeroes(self, nums: List[int]) -> None:
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slow = 0
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for fast in range(len(nums)):
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if nums[fast] != 0:
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nums[slow] = nums[fast]
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slow += 1
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for i in range(slow, len(nums)):
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nums[i] = 0
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```
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交换前后变量，避免补零
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```python
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def moveZeroes(self, nums: List[int]) -> None:
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slow, fast = 0, 0
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while fast < len(nums):
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if nums[fast] != 0:
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nums[slow], nums[fast] = nums[fast], nums[slow]
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slow += 1 # 保持[0, slow)区间是没有0的
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fast += 1
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```
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### Go：
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```go
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func moveZeroes(nums []int) {
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slow := 0
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for fast := 0; fast < len(nums); fast ++ {
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if nums[fast] != 0 {
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temp := nums[slow]
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nums[slow] = nums[fast]
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nums[fast] = temp
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slow++
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}
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}
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}
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```
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### JavaScript：
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```javascript
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var moveZeroes = function(nums) {
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let slow = 0;
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for(let fast = 0; fast < nums.length; fast++){
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if(nums[fast] != 0){//找到非0的元素
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nums[slow] = nums[fast];//把非0的元素赋值给数组慢指针指向的索引处的值
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slow++;//慢指针向右移动
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}
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}
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// 后面的元素全变成 0
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for(let j = slow; j < nums.length; j++){
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nums[j] = 0;
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}
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};
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```
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### TypeScript：
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```typescript
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function moveZeroes(nums: number[]): void {
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const length: number = nums.length;
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let slowIndex: number = 0,
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fastIndex: number = 0;
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while (fastIndex < length) {
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if (nums[fastIndex] !== 0) {
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nums[slowIndex++] = nums[fastIndex];
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};
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fastIndex++;
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}
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while (slowIndex < length) {
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nums[slowIndex++] = 0;
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}
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};
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```
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### C
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```c
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void moveZeroes(int* nums, int numsSize){
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int fastIndex = 0, slowIndex = 0;
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for (; fastIndex < numsSize; fastIndex++) {
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if (nums[fastIndex] != 0) {
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nums[slowIndex++] = nums[fastIndex];
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}
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}
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// 将slowIndex之后的元素变为0
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for (; slowIndex < numsSize; slowIndex++) {
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nums[slowIndex] = 0;
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}
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}
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```
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<p align="center">
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<a href="https://programmercarl.com/other/kstar.html" target="_blank">
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<img src="../pics/网站星球宣传海报.jpg" width="1000"/>
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</a>
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