LeetCode OJ：Maximal Square（最大矩形）

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 0
1 0 1 1 1
1 1 1 1 1
1 0 0 1 0

在给定的二维字符数组，找出最大全为1的矩形包含的1的个数：

注意这里相当于求面积而已，只要求出边长就很好办了。边长用dp很好求，这题其实跟前面有道dp问题很像，以后有时间找出来，先贴下代码：

 1 class Solution {
 2 public:
 3     int maximalSquare(vector<vector<char>>& matrix) {
 4         if(!matrix.size()||!matrix[0].size()) return 0;
 5         vector<vector<int>>dp(matrix.size(), vector<int>(matrix[0].size(), 0));
 6         int maxVal = 0;
 7         for(int i = 0; i < matrix.size(); ++i){
 8             if(matrix[i][0] == ‘1‘){
 9                 dp[i][0] = 1;
10                 maxVal = 1;
11             }else dp[i][0] = 0;
12         }
13         for(int j = 0; j < matrix[0].size(); ++j){
14             if(matrix[0][j] == ‘1‘){
15                 dp[0][j] = 1;
16                 maxVal = 1;
17             }else dp[0][j] = 0;
18         }
19         for(int i = 1; i < matrix.size(); ++i){
20             for(int j = 1; j < matrix[0].size(); ++j){
21                 if(matrix[i][j] == ‘1‘){
22                     dp[i][j] = min(dp[i-1][j], min(dp[i][j-1], dp[i-1][j-1])) + 1;
23                     maxVal = max(maxVal, dp[i][j]);
24                 }else
25                     dp[i][j] = 0;
26             }
27         }
28         return maxVal*maxVal;
29     }
30 };