Maximal Square

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

Return 4.

[思路]

dynamic programing. 以当前点(x,y) = '1' 为右下角的最大正方形的边长f(x,y) = min( f(x-1,y), f(x,y-1), f(x-1,y-1)) + 1.

递推公式已建立, dp就自然而然了.

public class Solution {
public int maximalSquare(char[][] matrix) {
if(matrix==null || matrix.length==0 || matrix[0].length==0) return 0;

int n = matrix.length;
int m = matrix[0].length;

int[][] d = new int
[m];
int max = 0;

for(int i=0; i<n; i++) {
if(matrix[i][0]=='1') {
d[i][0] = 1;
max = 1;
}
}

for(int j=0; j<m; j++) {
if(matrix[0][j]=='1') {
d[0][j] = 1;
max = 1;
}
}
for(int i=1; i<n; i++) {
for(int j=1; j<m; j++) {
if(matrix[i][j]=='0') d[i][j]=0;
else {
d[i][j] = Math.min( Math.min( d[i-1][j], d[i][j-1]), d[i-1][j-1] ) + 1;
max = Math.max(max, d[i][j]);
}
}
}
return max*max;
}
}