435 lines
12 KiB
Markdown
435 lines
12 KiB
Markdown
<p align="center">
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<a href="https://mp.weixin.qq.com/s/RsdcQ9umo09R6cfnwXZlrQ"><img src="https://img.shields.io/badge/PDF下载-代码随想录-blueviolet" alt=""></a>
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<a href="https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw"><img src="https://img.shields.io/badge/刷题-微信群-green" alt=""></a>
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<a href="https://space.bilibili.com/525438321"><img src="https://img.shields.io/badge/B站-代码随想录-orange" alt=""></a>
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<a href="https://mp.weixin.qq.com/s/QVF6upVMSbgvZy8lHZS3CQ"><img src="https://img.shields.io/badge/知识星球-代码随想录-blue" alt=""></a>
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</p>
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<p align="center"><strong>欢迎大家<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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## 59.螺旋矩阵II
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[力扣题目链接](https://leetcode-cn.com/problems/spiral-matrix-ii/)
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给定一个正整数 n，生成一个包含 1 到 n2 所有元素，且元素按顺时针顺序螺旋排列的正方形矩阵。
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示例:
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输入: 3
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输出:
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[
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[ 1, 2, 3 ],
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[ 8, 9, 4 ],
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[ 7, 6, 5 ]
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]
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## 思路
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这道题目可以说在面试中出现频率较高的题目，**本题并不涉及到什么算法，就是模拟过程，但却十分考察对代码的掌控能力。**
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要如何画出这个螺旋排列的正方形矩阵呢？
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相信很多同学刚开始做这种题目的时候，上来就是一波判断猛如虎。
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结果运行的时候各种问题，然后开始各种修修补补，最后发现改了这里哪里有问题，改了那里这里又跑不起来了。
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大家还记得我们在这篇文章[数组：每次遇到二分法，都是一看就会，一写就废](https://programmercarl.com/0704.二分查找.html)中讲解了二分法，提到如果要写出正确的二分法一定要坚持**循环不变量原则**。
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而求解本题依然是要坚持循环不变量原则。
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模拟顺时针画矩阵的过程:
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* 填充上行从左到右
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* 填充右列从上到下
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* 填充下行从右到左
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* 填充左列从下到上
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由外向内一圈一圈这么画下去。
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可以发现这里的边界条件非常多，在一个循环中，如此多的边界条件，如果不按照固定规则来遍历，那就是**一进循环深似海，从此offer是路人**。
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这里一圈下来，我们要画每四条边，这四条边怎么画，每画一条边都要坚持一致的左闭右开，或者左开又闭的原则，这样这一圈才能按照统一的规则画下来。
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那么我按照左闭右开的原则，来画一圈，大家看一下：
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这里每一种颜色，代表一条边，我们遍历的长度，可以看出每一个拐角处的处理规则，拐角处让给新的一条边来继续画。
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这也是坚持了每条边左闭右开的原则。
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一些同学做这道题目之所以一直写不好，代码越写越乱。
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就是因为在画每一条边的时候，一会左开又闭，一会左闭右闭，一会又来左闭右开，岂能不乱。
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代码如下，已经详细注释了每一步的目的，可以看出while循环里判断的情况是很多的，代码里处理的原则也是统一的左闭右开。
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整体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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vector<vector<int>> generateMatrix(int n) {
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vector<vector<int>> res(n, vector<int>(n, 0)); // 使用vector定义一个二维数组
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int startx = 0, starty = 0; // 定义每循环一个圈的起始位置
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int loop = n / 2; // 每个圈循环几次，例如n为奇数3，那么loop = 1 只是循环一圈，矩阵中间的值需要单独处理
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int mid = n / 2; // 矩阵中间的位置，例如：n为3， 中间的位置就是(1，1)，n为5，中间位置为(2, 2)
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int count = 1; // 用来给矩阵中每一个空格赋值
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int offset = 1; // 每一圈循环，需要控制每一条边遍历的长度
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int i,j;
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while (loop --) {
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i = startx;
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j = starty;
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// 下面开始的四个for就是模拟转了一圈
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// 模拟填充上行从左到右(左闭右开)
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for (j = starty; j < starty + n - offset; j++) {
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res[startx][j] = count++;
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}
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// 模拟填充右列从上到下(左闭右开)
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for (i = startx; i < startx + n - offset; i++) {
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res[i][j] = count++;
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}
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// 模拟填充下行从右到左(左闭右开)
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for (; j > starty; j--) {
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res[i][j] = count++;
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}
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// 模拟填充左列从下到上(左闭右开)
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for (; i > startx; i--) {
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res[i][j] = count++;
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}
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// 第二圈开始的时候，起始位置要各自加1， 例如：第一圈起始位置是(0, 0)，第二圈起始位置是(1, 1)
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startx++;
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starty++;
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// offset 控制每一圈里每一条边遍历的长度
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offset += 2;
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}
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// 如果n为奇数的话，需要单独给矩阵最中间的位置赋值
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if (n % 2) {
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res[mid][mid] = count;
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}
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return res;
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}
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};
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```
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## 类似题目
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* 54.螺旋矩阵
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* 剑指Offer 29.顺时针打印矩阵
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## 其他语言版本
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Java：
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```Java
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class Solution {
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public int[][] generateMatrix(int n) {
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int[][] res = new int[n][n];
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// 循环次数
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int loop = n / 2;
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// 定义每次循环起始位置
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int startX = 0;
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int startY = 0;
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// 定义偏移量
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int offset = 1;
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// 定义填充数字
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int count = 1;
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// 定义中间位置
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int mid = n / 2;
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while (loop > 0) {
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int i = startX;
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int j = startY;
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// 模拟上侧从左到右
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for (; j<startY + n -offset; ++j) {
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res[startX][j] = count++;
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}
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// 模拟右侧从上到下
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for (; i<startX + n -offset; ++i) {
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res[i][j] = count++;
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}
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// 模拟下侧从右到左
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for (; j > startY; j--) {
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res[i][j] = count++;
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}
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// 模拟左侧从下到上
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for (; i > startX; i--) {
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res[i][j] = count++;
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}
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loop--;
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startX += 1;
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startY += 1;
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offset += 2;
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}
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if (n % 2 == 1) {
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res[mid][mid] = count;
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}
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return res;
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}
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}
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```
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python:
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```python3
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class Solution:
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def generateMatrix(self, n: int) -> List[List[int]]:
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# 初始化要填充的正方形
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matrix = [[0] * n for _ in range(n)]
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left, right, up, down = 0, n - 1, 0, n - 1
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number = 1 # 要填充的数字
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while left < right and up < down:
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# 从左到右填充上边
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for x in range(left, right):
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matrix[up][x] = number
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number += 1
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# 从上到下填充右边
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for y in range(up, down):
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matrix[y][right] = number
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number += 1
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# 从右到左填充下边
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for x in range(right, left, -1):
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matrix[down][x] = number
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number += 1
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# 从下到上填充左边
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for y in range(down, up, -1):
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matrix[y][left] = number
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number += 1
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# 缩小要填充的范围
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left += 1
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right -= 1
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up += 1
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down -= 1
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# 如果阶数为奇数，额外填充一次中心
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if n % 2:
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matrix[n // 2][n // 2] = number
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return matrix
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```
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javaScript
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```js
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/**
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* @param {number} n
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* @return {number[][]}
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*/
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var generateMatrix = function(n) {
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// new Array(n).fill(new Array(n))
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// 使用fill --> 填充的是同一个数组地址
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const res = Array.from({length: n}).map(() => new Array(n));
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let loop = n >> 1, i = 0, //循环次数
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count = 1,
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startX = startY = 0; // 起始位置
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while(++i <= loop) {
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// 定义行列
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let row = startX, column = startY;
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// [ startY, n - i)
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while(column < n - i) {
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res[row][column++] = count++;
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}
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// [ startX, n - i)
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while(row < n - i) {
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res[row++][column] = count++;
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}
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// [n - i , startY)
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while(column > startY) {
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res[row][column--] = count++;
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}
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// [n - i , startX)
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while(row > startX) {
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res[row--][column] = count++;
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}
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startX = ++startY;
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}
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if(n & 1) {
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res[startX][startY] = count;
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}
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return res;
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};
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```
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Go:
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```go
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func generateMatrix(n int) [][]int {
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top, bottom := 0, n-1
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left, right := 0, n-1
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num := 1
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tar := n * n
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matrix := make([][]int, n)
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for i := 0; i < n; i++ {
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matrix[i] = make([]int, n)
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}
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for num <= tar {
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for i := left; i <= right; i++ {
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matrix[top][i] = num
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num++
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}
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top++
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for i := top; i <= bottom; i++ {
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matrix[i][right] = num
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num++
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}
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right--
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for i := right; i >= left; i-- {
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matrix[bottom][i] = num
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num++
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}
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bottom--
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for i := bottom; i >= top; i-- {
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matrix[i][left] = num
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num++
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}
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left++
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}
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return matrix
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}
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```
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Swift:
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```swift
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func generateMatrix(_ n: Int) -> [[Int]] {
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var result = [[Int]](repeating: [Int](repeating: 0, count: n), count: n)
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var startRow = 0
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var startColumn = 0
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var loopCount = n / 2
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let mid = n / 2
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var count = 1
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var offset = 1
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var row: Int
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var column: Int
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while loopCount > 0 {
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row = startRow
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column = startColumn
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for c in column ..< startColumn + n - offset {
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result[startRow][c] = count
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count += 1
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column += 1
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}
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for r in row ..< startRow + n - offset {
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result[r][column] = count
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count += 1
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row += 1
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}
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for _ in startColumn ..< column {
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result[row][column] = count
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count += 1
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column -= 1
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}
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for _ in startRow ..< row {
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result[row][column] = count
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count += 1
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row -= 1
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}
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startRow += 1
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startColumn += 1
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offset += 2
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loopCount -= 1
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}
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if (n % 2) != 0 {
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result[mid][mid] = count
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}
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return result
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}
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```
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Rust:
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```rust
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impl Solution {
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pub fn generate_matrix(n: i32) -> Vec<Vec<i32>> {
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let mut res = vec![vec![0; n as usize]; n as usize];
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let (mut startX, mut startY, mut offset): (usize, usize, usize) = (0, 0, 1);
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let mut loopIdx = n/2;
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let mid: usize = loopIdx as usize;
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let mut count = 1;
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let (mut i, mut j): (usize, usize) = (0, 0);
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while loopIdx > 0 {
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i = startX;
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j = startY;
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while j < (startY + (n as usize) - offset) {
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res[i][j] = count;
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count += 1;
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j += 1;
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}
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while i < (startX + (n as usize) - offset) {
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res[i][j] = count;
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count += 1;
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i += 1;
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}
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while j > startY {
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res[i][j] = count;
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count += 1;
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j -= 1;
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}
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while i > startX {
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res[i][j] = count;
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count += 1;
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i -= 1;
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}
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startX += 1;
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startY += 1;
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offset += 2;
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loopIdx -= 1;
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}
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if(n % 2 == 1) {
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res[mid][mid] = count;
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}
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res
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}
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}
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```
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-----------------------
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* 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
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* B站视频：[代码随想录](https://space.bilibili.com/525438321)
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* 知识星球：[代码随想录](https://mp.weixin.qq.com/s/QVF6upVMSbgvZy8lHZS3CQ)
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<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码.jpg width=450> </img></div>
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