LeetCode #221 - Maximal Square

题目描述：

Given a 2D binary matrix filled with 0's and 1's, find the largest square containing only 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.

采用动态规划，用dp[i][j]表示以matrix[i][j]为右下角的正方形的边长，如果matrix[i][j]=1，那么dp[i][j]=min(dp[i-1][j-1]，dp[i-1][j]，dp[i][j-1])+1，否则dp[i][j]=0，然后取最大的pow(dp[i][j]，2)即可。

class Solution {
public:
    int maximalSquare(vector<vector<char>>& matrix) {
        if(matrix.size()==0) return 0;
        if(matrix[0].size()==0) return 0;
        vector<int> row(matrix[0].size(),0);
        vector<vector<int>> dp(matrix.size(),row);
        for(int i=0;i<matrix.size();i++)
        {
            if(matrix[i][0]=='0') dp[i][0]==0;
            else dp[i][0]=1;
        }
        for(int j=0;j<matrix[0].size();j++)
        {
            if(matrix[0][j]=='0') dp[0][j]==0;
            else dp[0][j]=1;
        }
        
        for(int i=1;i<matrix.size();i++)
        {
            for(int j=1;j<matrix[0].size();j++)
            {
                if(matrix[i][j]=='1')
                dp[i][j]=min(min(dp[i-1][j],dp[i][j-1]),dp[i-1][j-1])+1;
                else dp[i][j]=0;
            }
        }
        int result=INT_MIN;
        for(int i=0;i<matrix.size();i++)
            for(int j=0;j<matrix[0].size();j++)
                result=max(result,dp[i][j]);
        return pow(result,2);
    }
};