Given n non-negative integers representing the histogram's bar height where the width of each bar is 1, find the area of largest rectangle in the histogram.
Above is a histogram where width of each bar is 1, given height =[2,1,5,6,2,3].
The largest rectangle is shown in the shaded area, which has area =10unit.
For example,
Given height =[2,1,5,6,2,3],
return10.
思路:
1、如果已知height数组是升序的,应该怎么做?
比如1,2,5,7,8
那么就是(1*5) vs. (2*4) vs. (5*3) vs. (7*2) vs. (8*1)
也就是max(height[i]*(size-i))
2、使用栈的目的就是构造这样的升序序列,按照以上方法求解。
但是height本身不一定是升序的,应该怎样构建栈?
比如2,1,5,6,2,3
(1)2进栈。s={2}, result = 0
(2)1比2小,不满足升序条件,因此将2弹出,并记录当前结果为2*1=2。
将2替换为1重新进栈。s={1,1}, result = 2
(3)5比1大,满足升序条件,进栈。s={1,1,5},result = 2
(4)6比5大,满足升序条件,进栈。s={1,1,5,6},result = 2
(5)2比6小,不满足升序条件,因此将6弹出,并记录当前结果为6*1=6。s={1,1,5},result = 6
2比5小,不满足升序条件,因此将5弹出,并记录当前结果为5*2=10(因为已经弹出的5,6是升序的)。s={1,1},result = 10
2比1大,将弹出的5,6替换为2重新进栈。s={1,1,2,2,2},result = 10
(6)3比2大,满足升序条件,进栈。s={1,1,2,2,2,3},result = 10
栈构建完成,满足升序条件,因此按照升序处理办法得到上述的max(height[i]*(size-i))=max{3*1, 2*2, 2*3, 2*4, 1*5, 1*6}=8<10
综上所述,result=10
1 class Solution { 2 public: 3 int largestRectangleArea(vector<int> &height) { 4 int n=height.size(); 5 if(n<1) return 0; 6 stack<int> s; 7 height.push_back(0); 8 int max=0; 9 for(int i=0;i<n+1;i++){ 10 while(!s.empty()&&height[s.top()]>=height[i]){ 11 int index=s.top(); 12 s.pop(); 13 int res=height[index]*(s.empty()?i:(i-s.top()-1)) ; 14 if(res>max) max=res; 15 } 16 s.push(i); 17 } 18 return max; 19 } 20 };