Leetcode | Maximal Rectangle

Given a 2D binary matrix filled with 0's and 1's, find the largest rectangle containing all ones and return its area.

穷举很难。之前做Largest Rectangle in Histogram的时候，看到网上有人说这两道题很类似。于是开始往这方面想。

思路如下：

1. 对于第j列，每一行可以算出从j这个位置开始，连续的'1'的个数。这样在第j列就可以计算得到一个histogram。

2. 然后直接应用Largest Rectangle in Histogram的O(n)算法求最大值；

3. 所有最大值再求最大，就是结果。

 1 class Solution {

 2 public:

 3     int largestRectangleArea(vector<int> &height) {

 4         int n = height.size();

 5         if (n == 0) return 0;

 6         

 7         int max = 0, last, area;

 8         

 9         stack<int> indexes;

10         height.push_back(0);

11         

12         for (int i = 0; i < height.size(); ) {

13             if (indexes.empty() || (height[i]>=height[indexes.top()])) {

14                 indexes.push(i);

15                 i++;

16             } else {

17                 last = height[indexes.top()];

18                 indexes.pop();

19                 area = last * (indexes.empty() ? i : i - indexes.top() - 1);

20                 if (area > max) max = area;

21             }

22         }

23         

24         return max;

25     }

26     int maximalRectangle(vector<vector<char> > &matrix) {

27         int m = matrix.size();

28         if (m == 0) return 0;

29         int n = matrix[0].size();

30         if (n == 0) return 0;

31         

32         vector<int> hist(m, 0);

33         int max = 0, area;

34         

35         for (int j = 0; j < n; ++j) {

36             for (int i = 0; i < m; ++i) {

37                 hist[i] = 0; 

38                 for (int k = j; k < n && matrix[i][k] == '1'; ++k) hist[i]++;  

39             }        

40             area = largestRectangleArea(hist);

41             if (area > max) max = area;

42         }

43         return max;

44     }

45 };

当然这样建histogram的复杂度是O(n*m*n)。

建historgram可以事先建好。动态规划啊，竟然没想到。。。。

 1 class Solution {

 2 public:

 3     int largestRectangleArea(vector<int> &height) {

 4         int n = height.size();

 5         if (n == 0) return 0;

 6         

 7         int max = 0, last, area;

 8         

 9         stack<int> indexes;

10         height.push_back(0);

11         

12         for (int i = 0; i < height.size(); ) {

13             if (indexes.empty() || (height[i]>=height[indexes.top()])) {

14                 indexes.push(i);

15                 i++;

16             } else {

17                 last = height[indexes.top()];

18                 indexes.pop();

19                 area = last * (indexes.empty() ? i : i - indexes.top() - 1);

20                 if (area > max) max = area;

21             }

22         }

23         

24         return max;

25     }

26     int maximalRectangle(vector<vector<char> > &matrix) {

27         int m = matrix.size();

28         if (m == 0) return 0;

29         int n = matrix[0].size();

30         if (n == 0) return 0;

31         

32         vector<vector<int> > hists(n + 1, vector<int>(m, 0));

33        

34         for (int j = n - 1; j >= 0; --j) {

35             for (int i = 0; i < m; ++i) {

36                 if (matrix[i][j] == '0') {

37                     hists[j][i] = 0;

38                 } else {

39                     hists[j][i] = hists[j + 1][i] + 1;

40                 }

41             }        

42         }

43         

44         int max = 0, area;    

45         for (int j = n - 1; j >= 0; --j) {

46             area = largestRectangleArea(hists[j]);

47             if (area > max) max = area;

48         }

49         

50         return max;

51     }

52 };

 这样就是O(m*n)了。空间复杂度是O(m*n)。