LeetCode-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 height = [2,1,5,6,2,3],
return 10.

一个栈，栈底的元素小于栈顶，每次压入新元素之前弹出所有大于新元素的就元素

所有点进栈一次，出栈时确认该点为最小值的作用域（起点为栈内上一个元素下标的后一位，终点为当前压入元素下标前一位） 复杂度O(n)

class Solution {
public:
    int largestRectangleArea(vector<int> &height) {
        if(height.size()==0)return 0;
        stack<pair<int,int> > sta;
        pair<int,int> p,q;
        int max=0;
        int start,end,area;
        for(int i=0;i<height.size();i++){
            p.first=height[i];
            p.second=i;
            if(sta.size()==0){
                sta.push(p);
            }
            else{
                while(sta.top().first>p.first){
                    q=sta.top();
                    sta.pop();
                    end=i-1;
                    if(sta.size()==0){
                        start=0;
                        area=q.first*(end-start+1);
                        if(area>max)max=area;
                        break;
                    }
                    else{
                        start=sta.top().second+1;
                        area=q.first*(end-start+1);
                        if(area>max)max=area;
                    }
                    
                }
                if(sta.size()!=0){
                    if(sta.top().first==p.first){
                        sta.top().second=i;
                    }
                    else{
                        sta.push(p);
                    }
                }
                else{
                    sta.push(p);
                }
            }
        }
        end=height.size()-1;
        while(sta.size()!=0){
            q=sta.top();
            sta.pop();
            if(sta.size()==0){
                start=0;
                area=q.first*(end-start+1);
                if(area>max)max=area;
            }
            else{
                start=sta.top().second+1;
                area=q.first*(end-start+1);
                if(area>max)max=area;
            }
        }
        return max;
    }
};