LeetCode Maximal Square

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

For example, given the following matrix:

1 0 1 0 01 0 1 1 11 1 1 1 11 0 0 1 0

Return 4.

思路分析：这题考察DP，缓存中间结果减少重复计算，DP方程为

if(matrix[i][j] != 0) dp[i,j] = min{dp[i-1,j], dp[i,j-1], dp[i-1,j-1]} + 1; else dp[i,j] = 0

其中dp[i,j]表示以[i,j]为右下角的区域内的最大的正方形的边长，最后返回dp数组中最大值的平方即为所求。

AC Code

public class Solution {    public int maximalSquare(char[][] matrix) {        //dp equation: if(matrix[i][j] != 0) dp[i,j] = min{dp[i-1,j], dp[i,j-1], dp[i-1,j-1]} + 1; else dp[i,j] = 0        int m = matrix.length;        if(m == 0) return 0;        int n = matrix[0].length;        //1227        int [][] dp = new int[m][n];        int max = 0;        for(int i = 0; i < m; i++){            if(matrix[i][0] == '1') {                dp[i][0] = 1;                max = Math.max(max, dp[i][0]);            }         }                for(int j = 0; j < n; j++){            if(matrix[0][j] == '1') {                dp[0][j] = 1;                max = Math.max(max, dp[0][j]);            }         }                for(int i = 1; i < m; i++){            for(int j = 1; j < n; j++){                if(matrix[i][j] == '1'){                    dp[i][j] =  Math.min(Math.min(dp[i-1][j], dp[i][j-1]), dp[i-1][j-1]) + 1;                    max = Math.max(max, dp[i][j]);                } else {                    dp[i][j] = 0;                }            }        }        return max * max;    }    //1234}

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