[LeetCode] 221. Maximal Square 最大正方形

Given a 2D binary matrix filled with 0's and 1's, find the largest square containing only 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. 

Credits:
Special thanks to @Freezen for adding this problem and creating all test cases.

给一个2维的01矩阵，找出最大只含有1的正方形，返回它的面积。

解法1: Brute force，对于矩阵中的每一个为1的点，都把它当作正方形的左上角，然后判断不同大小的正方形内的点是不是都为1。

解法2: DP，对于第一种的解法肯定有很多的重复计算，所以可以用DP的方法，把某一点能组成的最大正方形记录下来。

Python: DP

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class Solution:     # @param {character[][]} matrix     # @return {integer}     def maximalSquare(self, matrix):         if not matrix:             return 0           m, n = len(matrix), len(matrix[0])         size = [[0 for j in xrange(n)] for i in xrange(m)]         max_size = 0                   for j in xrange(n):             if matrix[0][j] == '1':                 size[0][j] = 1             max_size = max(max_size, size[0][j])                       for i in xrange(1, m):             if matrix[i][0] == '1':                 size[i][0] = 1             else:                 size[i][0] = 0             for j in xrange(1, n):                 if matrix[i][j] == '1':                     size[i][j] = min(size[i][j - 1],  \                                      size[i - 1][j],  \                                      size[i - 1][j - 1]) + 1                     max_size = max(max_size, size[i][j])                 else:                     size[i][j] = 0                               return max_size * max_size

C++:

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class Solution { public:     int maximalSquare(vector<vector<char>>& matrix) {         if (matrix.empty() || matrix[0].empty()) return 0;         int m = matrix.size(), n = matrix[0].size(), res = 0;         vector<vector<int>> dp(m, vector<int>(n, 0));         for (int i = 0; i < m; ++i) {             for (int j = 0; j < n; ++j) {                 if (i == 0 || j == 0) dp[i][j] = matrix[i][j] - '0';                 else if (matrix[i][j] == '1') {                     dp[i][j] = min(dp[i - 1][j - 1], min(dp[i][j - 1], dp[i - 1][j])) + 1;                 }                 res = max(res, dp[i][j]);             }         }         return res * res;     } };

C++:

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class Solution { public:     int maximalSquare(vector<vector<char>>& matrix) {         if (matrix.empty() || matrix[0].empty()) return 0;         int m = matrix.size(), n = matrix[0].size(), res = 0, pre = 0;         vector<int> dp(m + 1, 0);         for (int j = 0; j < n; ++j) {             for (int i = 1; i <= m; ++i) {                 int t = dp[i];                 if (matrix[i - 1][j] == '1') {                     dp[i] = min(dp[i], min(dp[i - 1], pre)) + 1;                     res = max(res, dp[i]);                 } else {                     dp[i] = 0;                 }                 pre = t;             }         }         return res * res;     } };

　

 

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