[LeetCode] 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:

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：

转化成：

 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 };