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 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

Return 4. 

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 0;

    int rows = matrix.size();//行数
    int cols = matrix[0].size(); //列数

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

    for (int i = 0; i < rows; i ++) {
        for (int j = 0; j < cols; j ++) {
            if (matrix[i][j] == '1') {
                int temp = min(dp[i][j], dp[i][j+1]);
                dp[i+1][j+1] = min(temp, dp[i+1][j]) + 1;
            }
            else 
                dp[i+1][j+1] = 0;
            
            result = max(result, dp[i+1][j+1]);
        }
    }
    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;
// }