221. Maximal Square(难)

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 01 0 1 1 11 1 1 1 11 0 0 1 0

Return 4.

构造一个新的矩阵dp，dp[i][j]表示以点(i, j)为右下角的正方形的边长；状态转移方程：

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

对于题目所给的例子就有：

1 0 1 0 01 0 1 1 11 1 1 1 11 0 0 1 0  

转化成：

1 0 1 0 01 0 1 1 11 1 1 2 11 0 0 1 0  

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