Maximal Square

Problem:

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.

Solutions:

定义p[i][j]为右下角是点matrix[i][j]的最大全'1'正方形，于是

　　若matrix[i][j] == 0,则p[i][j] = 0;

　　若i == 0,则p[0][j] = matrix[0][j] == '0' ? 0 : 1;

　　若j == 0,则p[i][0] = matrix[i][0] == '0' ? 0 : 1;

　　若i > 0 && j > 0 && matrix[i][j] == '1', 则p[i][j] = min(p[i-1][j], p[i][j-1], p[i-1][j-1]) + 1;

分析发现p[i][j]只依赖于上一行p[i-1][]和当前行的前一个元素p[i][j-1]，于是空间开销可以进一步压缩到O(列数），以下是实现代码（C++)：

int maximalSquare(vector<vector<char>>& matrix) {
int n, m;
if((n = matrix.size()) < 1 || (m = matrix[0].size()) < 1) return 0;
vector<int> dp(m, 0);
int pre, cur, res = 0;
for(int i = 0; i < n; ++i){
pre = matrix[i][0] - '0';
if(res < pre) res = pre;
for(int j = 1; j < m; ++j){
if(matrix[i][j] == '1'){
cur = min(pre, min(dp[j-1], dp[j])) + 1;
if(cur > res) res = cur;
}else{
cur = 0;
}
dp[j-1] = pre;
pre = cur;
}
dp[m-1] = pre;
}
return res * res;
}