221. Maximal Square - jiejackyzhang/leetcode-note GitHub Wiki

Given a 2D binary matrix filled with 0's and 1's, find the largest square containing only 1's and return its area.

For example, given the following matrix:

1 0 1 0 0
1 0 1 1 1
1 1 1 1 1
1 0 0 1 0

Return 4.

解题思路为Dynamic Programming。

令dp[i][j]表示以matrix[i+1][j+1]为右下角的max square的边长。 左，上，和左上三个格子决定了square的大小，当前格子若是1的话，这个格子max square的边长为以上三者的最小值+1。

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