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.
Credits:
Special thanks to @Freezen for adding this problem and creating all test cases.
思路借鉴：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.
代码如下：
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[n][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;
}
}
运行结果：
