【数据结构基础

原题连接：Leetcode 59. Spiral Matrix II

Given a positive integer n, generate an n x n matrix filled with elements from 1 to n² in spiral order.

Example 1:

Input: n = 3
Output: [[1,2,3],[8,9,4],[7,6,5]]

Example 2:

Input: n = 1
Output: [[1]]

Constraints:

• 1 <= n <= 20

模拟呗，设置四个方向
每次判断下一格是否越界，越界就转向

class Solution {
public:
    vector> generateMatrix(int n) {
        // maxn最大值 cur当前值
        int maxn = n * n, cur = 1;

        // 方向: 右下左上
        vector> direction = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
        vector> ans(n, vector(n));

        // xy坐标 i方向的下标
        int x = 0, y = 0, i = 0;;

        while(cur <= maxn){
            ans[x][y] = cur;
            cur++;

            // 若果下一个方向越界 或者碰到了已经填过数的格子 将方向转向
            int next_x = x + direction[i][0], next_y = y + direction[i][1];
            if(next_x < 0 || next_x >= n || next_y < 0 || next_y >= n || ans[next_x][next_y] != 0){
                i = (i + 1) % 4;
            }

            x = x + direction[i][0];
            y = y + direction[i][1];
        }

        return ans;
    }
};

• 时间复杂度：O(n²)，循环进行n²次
• 空间复杂度：O(1)，常数个临时变量