[LeetCode]Maximal Square

最大全1平方矩阵面积

最新推荐文章于 2020-04-25 23:48:14 发布

原创最新推荐文章于 2020-04-25 23:48:14 发布 · 430 阅读

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本文介绍了一种求解二维二进制矩阵中最大全1正方形的方法，通过动态规划构造辅助矩阵Dp[m*n]，并给出了具体的动态方程及C++实现代码。

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:

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;
    }
};