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

Return 4.

枚举就OK了，时间复杂度为O(n^3)，276ms AC。

1 class Solution { 2 public: 3     int getArea(vector
     
       &array, int k) { 4         if (array.size() < k) return 0; 5         int cnt = 0; 6         for (int i = 0; i < array.size(); ++i) { 7             if (array[i] != k) cnt = 0; 8             else ++cnt; 9             if (cnt == k) return k * k;10         }11         return 0;12     }13     int maximalSquare(vector
      
       
        >& matrix) {14         vector
        
          array;15         int res = 0;16         for (int i = 0; i < matrix.size(); ++i) {17             array.assign(matrix[0].size(), 0);18             for (int j = i; j < matrix.size(); ++j) {19                 for (int k = 0; k < matrix[0].size(); ++k) if (matrix[j][k] == '1') ++array[k];20                 res = max(res, getArea(array, j-i+1));21             }22         }23         return res;24     }25 };
        
       
      
     

 

可以使用动态规划优化到O(n^2)，下面的代码12ms AC。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;

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

1 class Solution { 2 public: 3     int maximalSquare(vector
     
      
       >& matrix) { 4         if (matrix.empty() || matrix[0].empty()) return 0; 5         int M = matrix.size(), N = matrix[0].size(), res = 0; 6         vector
       
        
         > dp(M, vector
         
          (N, 0)); 7         for (int i = 0; i < M; ++i) if (matrix[i][0] == '1') { 8             dp[i][0] = 1; res = 1; 9         }10         for (int j = 0; j < N; ++j) if (matrix[0][j] == '1') {11             dp[0][j] = 1; res = 1;12         }13         for (int i = 1; i < M; ++i) {14             for (int j = 1; j < N; ++j) {15                 if (matrix[i][j] == '1') 16                     dp[i][j] = min(dp[i-1][j-1], min(dp[i-1][j], dp[i][j-1])) + 1;17                 res = max(res, dp[i][j]);18             }19         }20         return res * res;21     }22 };