[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 01 0 1 1 11 1 1 1 11 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][n] ( 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;    }};