題目：
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.
解題思路：
這種包含最大、最小等含優化的字眼時，一般都需要用到動態規劃進行求解。本題求面積我們可以轉化為求邊長，由於是正方形，因此可以根據正方形的四個角的座標寫出動態規劃的轉移方程式（畫一個圖，從左上角推到右下角，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
根據此方程，就可以寫出如下的程式碼：
程式碼展示：
1 #include <iostream> 2 #include <vector> 3 #include <cstring> 4 using namespace std; 5 6 //dp[i][j] = min(dp[i-1][j-1], dp[i-1][j], dp[i][j-1]) + 1; 7 //where matrix[i][j] == 1 8 int maximalSquare(vector<vector<char>>& matrix) 9 { 10 if (matrix.empty()) 11 return 0; 12 13 int rows = matrix.size();//行數 14 int cols = matrix[0].size(); //列數 15 16 vector<vector<int> > dp(rows+1, vector<int>(cols+1, 0)); 17 /* 18 0 0 0 0 0 0 19 0 1 0 1 0 0 20 0 1 0 1 1 1 21 0 1 1 1 1 1 22 0 1 0 0 1 0 23 */ 24 int result = 0; //return result 25 26 for (int i = 0; i < rows; i ++) { 27 for (int j = 0; j < cols; j ++) { 28 if (matrix[i][j] == '1') { 29 int temp = min(dp[i][j], dp[i][j+1]); 30 dp[i+1][j+1] = min(temp, dp[i+1][j]) + 1; 31 } 32 else 33 dp[i+1][j+1] = 0; 34 35 result = max(result, dp[i+1][j+1]); 36 } 37 } 38 return result * result; //get the area of square; 39 } 40 41 // int main() 42 // { 43 // char *ch1 = "00000"; 44 // char *ch2 = "00000"; 45 // char *ch3 = "00000"; 46 // char *ch4 = "00000"; 47 // vector<char> veccol1(ch1, ch1 + strlen(ch1)); 48 // vector<char> veccol2(ch2, ch2 + strlen(ch2)); 49 // vector<char> veccol3(ch3, ch3 + strlen(ch3)); 50 // vector<char> veccol4(ch4, ch4 + strlen(ch4)); 51 // 52 // vector<vector<char> > vecrow; 53 // vecrow.push_back(veccol1); 54 // vecrow.push_back(veccol2); 55 // vecrow.push_back(veccol3); 56 // vecrow.push_back(veccol4); 57 // 58 // vector<vector<char> > vec; 59 // cout << maximalSquare(vec); 60 // return 0; 61 // }