Leetcode 0221: Maximal Square

题目描述：

Given an m x n binary matrix filled with 0’s and 1’s, find the largest square containing only 1’s and return its area.

Example 1:

Input: matrix = [[“1”,“0”,“1”,“0”,“0”],[“1”,“0”,“1”,“1”,“1”],[“1”,“1”,“1”,“1”,“1”],[“1”,“0”,“0”,“1”,“0”]]
Output: 4

Example 2:

Input: matrix = [[“0”,“1”],[“1”,“0”]]
Output: 1

Example 3:

Input: matrix = [[“0”]]
Output: 0

Constraints:

m == matrix.length
n == matrix[i].length
1 <= m, n <= 300
matrix[i][j] is ‘0’ or ‘1’.

Time complexity: O(nm)
动态规划：
状态表示：dp[i][j]表示以(i,j)为右下角的最大正方形的边长
状态转移：
以(i,j)为右下角的最大正方形的边长取决于其相邻的 上边(i-1,j)，左边(i,j-1)，左上(i-1,j-1) 此三个点的最大正方形的边长, 取三者的最小值+1

class Solution {
    
    
    public int maximalSquare(char[][] matrix) {
    
    
        int m = matrix.length;
        int n = matrix[0].length;
        int[][] dp = new int[m+1][n+1];
        int res = 0;
        for(int i = 1; i <= m; i++){
    
    
            for(int j = 1; j <= n; j++){
    
    
                if(matrix[i-1][j-1] == '1'){
    
    
                    dp[i][j] = Math.min(Math.min(dp[i][j - 1], dp[i - 1][j]), dp[i - 1][j - 1]) + 1;
                    res = Math.max(res,  dp[i][j]);
                }
            }
        }
        return res*res;
    }
}