LeetCode: 221_Maximal Square | 二维0-1矩阵中计算包含1的最大正方形的面积 | Medium

题目：

Given a 2D binary matrix filled with 's and 1's, find the largest square containing all 's and return its area. 

For example, given the following matrix: 

Return . 

解题思路：

　　这种包含最大、最小等含优化的字眼时，一般都需要用到动态规划进行求解。本题求面积我们可以转化为求边长，由于是正方形，因此可以根据正方形的四个角的坐标写出动态规划的转移方程式（画一个图，从左上角推到右下角，很容易理解）：

dp[i][j] = min(dp[i-1][j-1], dp[i-1][j], dp[i][j-1]) + 1; where matrix[i][j] == 1

根据此方程，就可以写出如下的代码：

代码展示：

 #include <iostream>
 #include <vector>
 #include <cstring>
 using namespace std;

 //dp[i][j] = min(dp[i-1][j-1], dp[i-1][j], dp[i][j-1]) + 1;
 //where matrix[i][j] == 1
 int maximalSquare(vector<vector<char>>& matrix)
 {
     if (matrix.empty())
         return ;

     int rows = matrix.size();//行数
     int cols = matrix[].size(); //列数

     vector<vector<int> > dp(rows+, vector<int>(cols+, ));
     /*
         0 0 0 0 0 0
         0 1 0 1 0 0
         0 1 0 1 1 1
         0 1 1 1 1 1
         0 1 0 0 1 0
     */
     int result = ; //return result

     for (int i = ; i < rows; i ++) {
         for (int j = ; j < cols; j ++) {
             if (matrix[i][j] == '') {
                 int temp = min(dp[i][j], dp[i][j+]);
                 dp[i+][j+] = min(temp, dp[i+][j]) + ;
             }
             else
                 dp[i+][j+] = ;

             result = max(result, dp[i+][j+]);
         }
     }
     return result * result; //get the area of square;
 }

 // int main()
 // {
 //     char *ch1 = "00000";
 //     char *ch2 = "00000";
 //     char *ch3 = "00000";
 //     char *ch4 = "00000";
 //     vector<char> veccol1(ch1, ch1 + strlen(ch1));
 //     vector<char> veccol2(ch2, ch2 + strlen(ch2));
 //     vector<char> veccol3(ch3, ch3 + strlen(ch3));
 //     vector<char> veccol4(ch4, ch4 + strlen(ch4));
 //
 //     vector<vector<char> > vecrow;
 //     vecrow.push_back(veccol1);
 //     vecrow.push_back(veccol2);
 //     vecrow.push_back(veccol3);
 //     vecrow.push_back(veccol4);
 //
 //     vector<vector<char> > vec;
 //     cout << maximalSquare(vec);
 //     return 0;
 // }