LeetCode:Maximus Square(DP)

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.

   dp[i][j]表示以坐标（i,j）为右下角的形成的正方形的最大边长。

 dp[i][j]=min(dp[i-1][j],min(dp[i-1][j-1],dp[i][j-1]))+1;

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