【leetcode】Maximal Rectangle

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.

 

 

使用dpHeight[]数组来记录到第i行为止，第j个位置垂直连续包含多少个1（包括matrxi[i][j]）。如：

1 0 1 1 0

1 1 0 1 0

0 1 1 1 1

有如下结果：

第1行： dpHeight[] = {1, 0, 1, 1, 0}

第2行： dpHeight[] = {2, 1, 0, 2, 0}

第3行： dpHeight[] = {0, 2, 1, 3, 0}

 

对每个dpHeight求最大矩形面积，利用Largest Rectangle in Histogram里面的求解方法即可

 

  

 

 1 class Solution {

 2 public:

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

 4        

 5        

 6         int m=matrix.size();

 7         if(m==0) return 0;

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

 9        

10         vector<int> dpHeight(n,0);

11        

12         int result=0;

13         for(int i=0;i<m;i++)

14         {

15             for(int j=0;j<n;j++)

16             {

17                 if(matrix[i][j]=='1') dpHeight[j]++;

18                 else dpHeight[j]=0;

19             }

20            

21             int tmp=lineMaximalRectangle(dpHeight);

22             if(tmp>result) result=tmp;

23         }

24        

25         return result;

26     }

27    

28     struct Node

29     {

30         int index;

31         int height;

32         Node(int i,int h):index(i),height(h){};

33     };

34    

35     int lineMaximalRectangle(vector<int> &dpHeight)

36     {

37        

38        

39         int len=dpHeight.size();

40         if(len==0) return 0;

41        

42         stack<Node> stk;

43         Node n(0,dpHeight[0]);

44         stk.push(n);

45        

46         int result=0;

47         for(int i=1;i<len;i++)

48         {

49             int height=dpHeight[i];

50            

51             if(height>=stk.top().height)

52             {

53                 stk.push(Node(i,height));

54             }

55             else

56             {

57                 int nodeIndex=i;

58                 while(!stk.empty()&&height<stk.top().height)

59                 {

60                     int tmp=(i-stk.top().index)*stk.top().height;

61                     if(tmp>result) result=tmp;

62                     nodeIndex=stk.top().index;

63                     stk.pop();

64                 }

65                 stk.push(Node(nodeIndex,height));

66             }

67         }

68         while(!stk.empty())

69         {

70             int index=stk.top().index;

71             int height=stk.top().height;

72             int tmp=(len-index)*height;

73             if(tmp>result) result=tmp;

74             stk.pop();

75         }

76        

77         return result;

78        

79     }

80 };