分析

• 本题的考点：偏移量技巧。

• 如果当前位置是(x, y)，则该点的上右下左的坐标分别为：(x-1, y), (x, y+1), (x+1, y), (x, y-1)。偏移量为(-1, 0), (0, 1), (1, 0), (0, -1)。

代码

• C++

class Solution {
public:
    vector<vector<int>> generateMatrix(int n) {

        vector<vector<int>> res(n, vector<int>(n));

        int dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0};
        for (int i = 1, x = 0, y = 0, d = 0; i <= n * n; i++) {
            res[x][y] = i;
            int a = x + dx[d], b = y + dy[d];
            if (a < 0 || a >= n || b < 0 || b >= n || res[a][b]) {
                d = (d + 1) % 4;
                a = x + dx[d], b = y + dy[d];
            }
            x = a, y = b;
        }
        return res;
    }
};

• Java

public class Solution {
    public int[][] generateMatrix(int n) {

        int[] dx = {-1, 0, 1, 0}, dy = {0, 1, 0, -1};
        int[][] res = new int[n][n];
        int x = 0, y = 0, d = 1;
        for (int i = 1; i <= n * n; i++) {
            res[x][y] = i;
            int a = x + dx[d], b = y + dy[d];
            if (a < 0 || a >= n || b < 0 || b >= n || res[a][b] != 0) {
                d = (d + 1) % 4;
                a = x + dx[d];
                b = y + dy[d];
            }
            x = a;
            y = b;
        }
        return res;
    }
}

时空复杂度分析

• 时间复杂度：$O(n^2)$。

• 空间复杂度：考虑存储结果的数组 $O(n^2)$。