221. Maximal Square

221 . Maximal Square

Difficulty: Medium

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.

解题思路

一道dp问题。其实我不是特别会做，但是基本思路就是如果当前位置是1，则计算它右下方三个点的最小的那个加一即可。也即是f(x,y) = min( f(x-1,y), f(x,y-1), f(x-1,y-1))+1。x，y为横纵坐标。

具体实现

class Solution {public:    int min(int i,int j)    {        if(i > j) return j;        return i;    }    int maxx(int i, int j)    {        if(i > j) return i;        return j;    }    int maximalSquare(vector<vector<char>>& matrix) {        if(matrix.size() == 0 || matrix[0].size() == 0)            return 0;        int n = matrix.size();        int m = matrix[0].size();        int arr[n][m];        int max = 0;        for(int i = 0;i < n;i++)        {            if(matrix[i][0] == '1')            {                arr[i][0] = 1;                max = 1;            }            else arr[i][0] = 0;        }        for(int i = 0;i < m;i++)        {            if(matrix[0][i] == '1')            {                arr[0][i] = 1;                max = 1;            }            else arr[0][i] = 0;        }        for(int i = 1;i < n;i++)        {            for(int j = 1;j < m;j++)            {                if(matrix[i][j] == '0') arr[i][j] = 0;                else                {                    arr[i][j] = min(min(arr[i-1][j],arr[i][j-1]),arr[i-1][j-1]) + 1;                    max = maxx(arr[i][j],max);                }            }        }        return max*max;    }};