221. Maximal Square

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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.

题意：给出一个矩阵，用字符0,1填充，求1组成的最大的方格。

思路：画一个二维数组，遇到0对应值写0，遇到1，判断对角线以及左边和上边的值，如果均不为0，则该值等于周围最小的值加1。用dp[m][n]记录点(m, n)时，以点(m ,n)为右下角的方格的边长值，则以该点为右下角点的方格的边长值为dp[m][n] = min{dp[m-1][n-1], dp[m-1][n], dp[m][n-1]}+1。

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

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