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class Solution {
    public int[][] generateMatrix(int n) {
        int totalNum = n * n;
        int count = 1;
        int[][] ans = new int[n][n];
        int left = 0, right = n - 1, up = 0, down = n - 1;
        while (count <= totalNum) {
            for (int i = left; i <= right; i++) {
                ans[up][i] = count++;
            }
            for (int i = up + 1; i <= down; i++) {
                ans[i][right] = count++;
            }
            for (int i = right - 1; i >= left && count <= totalNum; i--) {
                ans[down][i] = count++;
            }
            for (int i = down - 1; i > up && count <= totalNum; i--) {
                ans[i][left] = count++;
            }
            left += 1;
            right -= 1;
            up += 1;
            down -= 1;
        }
        return ans;
    }
}

spiral matrix的经典解法, 判断是否把所有格子照顾到, 因为我们知道格子的总数.
时间复杂度: O(n^2)
空间复杂度: O(1)