[LeetCode]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.

We need to construct a matrix Dp[m*n]. f(m,n) equals the situation when the square's right-down top point is(m,n) ,the square's large. So the Dynamic Equation is :

f(m.n) = matrix[m]
( m=0 or n=0 );

when matrix(m,n) = 1:f(m,n) = min(f(m-1,n),f(n-1,m),f(n-1,m-1))+1;

when matrix(m,n)= 0; f(m,n) = 0;

and search all the point,find the max square. It's the max square.(you can search it just when you construct f(m,n) matrix.)

class Solution {
public:
int maximalSquare(vector<vector<char>>& matrix) {
if(matrix.size()==0)
return 0;
vector<vector<int>> dp(matrix.size(),vector<int>(matrix[0].size()));
int res = dp[0][0];
for(int i=0;i<matrix.size();++i){
dp[i][0] = matrix[i][0]-'0';
if(dp[i][0]>res)
res = dp[i][0];
}
for(int j=0;j<matrix[0].size();++j){
dp[0][j] = matrix[0][j]-'0';
if(dp[0][j]>res)
res = dp[0][j];
}
for(int i=1;i<matrix.size();++i){
for(int j=1;j<matrix[0].size();++j){
if(matrix[i][j]=='0')
dp[i][j] = 0;
if(matrix[i][j]=='1')
{
dp[i][j] = min(dp[i-1][j],dp[i][j-1],dp[i-1][j-1])+1;
if(dp[i][j]>res)
res = dp[i][j];
}
}
}
return res*res;
}
int min(int a,int b,int c){
if(a<=b&&a<=c)
return a;
if(b<=a&&b<=c)
return b;
if(c<=a&&c<=b)
return c;
}
};