[LeetCode] Maximal Square

Maximal Square

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.

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<int> &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<vector<char>>& matrix) {

14         vector<int> 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 0

1 0 1 1 1

1 1 1 1 1

1 0 0 1 0  

转化成：

1 0 1 0 0

1 0 1 1 1

1 1 1 2 1

1 0 0 1 0  

 1 class Solution {

 2 public:

 3     int maximalSquare(vector<vector<char>>& matrix) {

 4         if (matrix.empty() || matrix[0].empty()) return 0;

 5         int M = matrix.size(), N = matrix[0].size(), res = 0;

 6         vector<vector<int>> dp(M, vector<int>(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 };