May LeetCoding Challenge21 之 动态规划的min使用

动态转移方程为：matrix[i][j] = Math.min(Math.min(matrix[i-1][j-1], matrix[i][j-1]), matrix[i-1][j]) + 1;

JAVA

class Solution {
    public int countSquares(int[][] matrix) {
        int r = matrix.length;
        int c = matrix[0].length;
        if(r == 0 || c == 0) return 0;
        int res = 0;
        for(int i = 1; i < r; i++){
            for(int j = 1; j < c; j++){
                if(matrix[i][j] != 0)
                    matrix[i][j] = Math.min(Math.min(matrix[i-1][j-1], matrix[i][j-1]), matrix[i-1][j]) + 1;
            }                 //关键点在于Math.min的使用
        }
        for(int i = 0; i < r; i++){
            for(int j = 0; j < c; j++){
                res += matrix[i][j];
            }
        }
        return res;
    }
}

 

Python3

class Solution:
    def countSquares(self, matrix: List[List[int]]) -> int:
        r = len(matrix)
        c = len(matrix[0]);
        res = 0
        for i in range(r):
            for j in range(c):
                if matrix[i][j] > 0 and i > 0 and j > 0:
                    matrix[i][j] = min(matrix[i-1][j-1], matrix[i-1][j], matrix[i][j-1])+1
                res += matrix[i][j]
        return res