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.

給一個2維的01矩陣，找出最大只含有1的正方形，返回它的面積。

解法1: Brute force，對於矩陣中的每一個為1的點，都把它當作正方形的左上角，然後判斷不同大小的正方形內的點是不是都為1。

解法2: DP，對於第一種的解法肯定有很多的重復計算，所以可以用DP的方法，把某一點能組成的最大正方形記錄下來。

Python: DP

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++:

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++:

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;
    }
};

[LeetCode] 221. Maximal Square 最大正方形