750. 角矩形的数量（动态规划）

class Solution {
    
    public int countCornerRectangles(int[][] grid) {
        if(grid.length < 2 || grid[0].length < 2) return 0;
        int m = grid.length, n = grid[0].length, res = 0;
        int[][] dp = new int[n][n]; // dp[i][j] 表示这一行[i,j]两个点都是1且向上数存在也是两个1的对数

        for(int i = 0; i < m; i++) {
            for(int j = 0; j < n; j++) {
                if(grid[i][j] == 0) continue;
                for(int k = j + 1; k < n; k++) {
                    if(grid[i][k] == 1) {
                        res += dp[j][k];
                        dp[j][k]++;
                    }
                }
            }
        }

        return res;
    }
}