1.题目描述

给你一个二维字符矩阵 grid，其中 grid[i][j] 可能是 'X'、'Y' 或 '.'，返回满足以下条件的子矩阵数量：

• 包含 grid[0][0]
• 'X' 和 'Y' 的频数相等。
• 至少包含一个 'X'。

示例 1：

输入： grid = [["X","Y","."],["Y",".","."]]

输出： 3

解释：

示例 2：

输入： grid = [["X","X"],["X","Y"]]

输出： 0

解释：

不存在满足 'X' 和 'Y' 频数相等的子矩阵。

2.解题思路

拿到这题首先可以想到使用dp数组存储以每一个点为结束位置的子矩阵中的X,Y个数，可以看出对于下标[i,j]，假设i,j均>=1,那么dp[i][j]对应矩阵的X，Y个数是可以通过dp[i-1][j],dp[i][j-1]以及dp[i-1][j-1]这三个子矩阵进行加减操作得到的。

因此，为了表示每个以i，j为结束位置的子矩阵中"X"和“Y”各自的个数，我们定义一个三维dp数组，dp[i][j][0]用于记录"X"的个数，dp[i][j][1]用于记录“Y”的个数，接下来只需要先初始化第0行和第0列这两种特殊情况，然后一般情况就可以通过递推式进行判断

3.代码实现

Java版

class Solution {public int numberOfSubmatrices(char[][] grid) {int m = grid.length, n = grid[0].length;int[][][] dp = new int[m][n][2];if (grid[0][0] == 'X') {dp[0][0][0] += 1;} else if (grid[0][0] == 'Y') {dp[0][0][1] += 1;}int ans = 0;//初始化第0行for (int j = 1; j < n; j++) {int x = dp[0][j-1][0];int y = dp[0][j-1][1];if (grid[0][j] == 'X') {x += 1;} else if (grid[0][j] == 'Y') {y += 1;}dp[0][j][0] = x;dp[0][j][1] = y;if (dp[0][j][0] >= 1 && dp[0][j][0] == dp[0][j][1]) {ans += 1;}}//初始化第0列for (int i = 1; i < m; i++) {int x = dp[i-1][0][0];int y = dp[i-1][0][1];if (grid[i][0] == 'X') {x += 1;} else if (grid[i][0] == 'Y') {y += 1;}dp[i][0][0] = x;dp[i][0][1] = y;if (dp[i][0][0] >= 1 && dp[i][0][0] == dp[i][0][1]) {ans += 1;}}//遍历for (int i = 1; i < m; i++) {for (int j = 1; j < n; j++) {int x = dp[i-1][j][0] + dp[i][j-1][0] - dp[i-1][j-1][0];int y = dp[i-1][j][1] + dp[i][j-1][1] - dp[i-1][j-1][1];if (grid[i][j] == 'X') {x += 1;} else if (grid[i][j] == 'Y') {y += 1;}dp[i][j][0] = x;dp[i][j][1] = y;if (dp[i][j][0] >= 1 && dp[i][j][0] == dp[i][j][1]) {ans += 1;}}}return ans;}
}

Python版

class Solution:def numberOfSubmatrices(self, grid: List[List[str]]) -> int:m,n = len(grid),len(grid[0])ans = 0dp = [[[0,0] for _ in range(n)] for _ in range(m)]if grid[0][0] == 'X':dp[0][0] = [1,0]elif grid[0][0] == 'Y':dp[0][0] = [0,1]#初始化第0行for j in range(1,n):if grid[0][j] == 'X':dp[0][j] = [dp[0][j-1][0] + 1,dp[0][j-1][1]]elif grid[0][j] == 'Y':dp[0][j] = [dp[0][j-1][0],dp[0][j-1][1] + 1]else:dp[0][j] = dp[0][j-1] if dp[0][j][0] >= 1 and dp[0][j][0] == dp[0][j][1]:ans += 1#初始化第0列for i in range(1,m):if grid[i][0] == 'X':dp[i][0] = [dp[i-1][0][0] + 1,dp[i-1][0][1]]elif grid[i][0] == 'Y':dp[i][0] = [dp[i-1][0][0],dp[i-1][0][1] + 1]else:dp[i][0] = dp[i-1][0]if dp[i][0][0] >= 1 and dp[i][0][0] == dp[i][0][1]:ans += 1for i in range(1,m):for j in range(1,n):x = dp[i-1][j][0] + dp[i][j-1][0] - dp[i-1][j-1][0]y = dp[i-1][j][1] + dp[i][j-1][1] - dp[i-1][j-1][1]if grid[i][j] == 'X':dp[i][j] = [x+1,y]elif grid[i][j] == 'Y':dp[i][j] = [x,y+1]else:dp[i][j] = [x,y]if dp[i][j][0] >= 1 and dp[i][j][0] == dp[i][j][1]:ans += 1return ans