Leetcode-84. Largest Rectangle in Histogram
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 = 10
unit.
For example,
Given heights = [2,1,5,6,2,3]
,
return 10
.
思路:
使用Stack。
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
code:
1 class Solution { 2 public: 3 int largestRectangleArea(vector<int>& heights) { 4 int result=0; 5 stack<int> silo; 6 for(int i=0;i<heights.size();i++) 7 { 8 if(silo.empty()||silo.top()<=heights[i]) 9 { 10 silo.push(heights[i]); 11 }else 12 { 13 int count=0; 14 while(!silo.empty()&&heights[i]<silo.top()) 15 { 16 count++; 17 result=max(result,silo.top()*count); 18 silo.pop(); 19 } 20 while(count--) 21 silo.push(heights[i]); 22 silo.push(heights[i]); 23 } 24 } 25 int count=1; 26 while(!silo.empty()) 27 { 28 result=max(result,silo.top()*count); 29 count++; 30 silo.pop(); 31 } 32 return result; 33 } 34 };