题目描述

题解思路

使用一个全局变量current记录当前遍历到的元素的值

我们只需要一圈一圈的从外向内遍历矩阵，每一圈遍历顺序为上边、右边、下边、左边，每遍历完一个元素后current++

我们需要注意的是如果上边与下边重合或者是右边与左边重合，我们只需要遍历上边、右边即可

题解代码

impl Solution {pub fn generate_matrix(n: i32) -> Vec<Vec<i32>> {let mut ans = vec![vec![0; n as usize]; n as usize];let (mut t, mut b, mut l, mut r) = (0usize, n as usize - 1, 0usize, n as usize - 1);let mut current = 1;// 从外圈向内圈遍历while l <= r && t <= b {// 上边 从左到右for i in l..=r {ans[t][i] = current;current += 1;}// 右边 从上到下for i in (t + 1)..=b {ans[i][r] = current;current += 1;}if l < r && t < b {// 下边 从右到左for i in ((l + 1)..r).rev() {ans[b][i] = current;current += 1;}// 左边 从下到上for i in ((t + 1)..=b).rev() {ans[i][l] = current;current += 1;}}l += 1;if r != 0 {r -= 1;}t += 1;if b != 0 {b -= 1;}}ans}
}

题目链接

https://leetcode.cn/problems/spiral-matrix-ii/