单调栈专题

问题驱动型分析博客：

1.

题目是这样的，给一个数组，返回一个大小相同的数组。返回的数组的第i个位置的值应当是，对于原数组中的第i个元素，至少往右走多少步，才能遇到一个比自己大的元素（如果之后没有比自己大的元素，或者已经是最后一个元素，则在返回数组的对应位置放上-1）。

简单的例子：

input: 5,3,1,2,4

return: -1 3 1 1 -1

 BONUS：

1.add 0 is important action to make sure alll elements can out of stack

2.when we see a situation we will have a static situation in our mind.But every time different situation's variable have diifferent meaning .

ONCE AGAIN:

MAKE SURE YOU understand the meaning of the variables!!!

#include<stdio.h>
#include<iostream>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<algorithm>
#include<stack>
#include<climits>
//monotonic stack application
using namespace std;
class Solution {
public:
    int largestRectangleArea(vector<int> &height)
    {
        int res=0;
        //add 0 make sure all elements come out of the stack
        height.push_back(0);
        vector<int> index;
        for(int i=0;i<height.size();i++)
        {
            //find the max area height need to find the lowest one,if larger some part can't make this rectangle
            while(index.size()>0&&height[index.back()]>=height[i])
            {
                int h=height[index.back()];
                index.pop_back();
                //if still have number in stack.Means rectangle don't start with zero,-1 is to balance
                int sidx=index.size()>0?index.back():-1;
                cout<<"indexback"<<index.back()<<endl;
                cout<<"flag"<<i<<" "<<sidx<<" "<<i-sidx-1<<endl;
                //make sure the true meaning of variable is very important.
                /**
                *such as this i is the point+1,sidx is the point0-1,point-point0+1==point+1-point0+1-1
                point0&point means the rectangle's start and end
                **/
                res=max(res,h*(i-sidx-1));

            }
            index.push_back(i);

        }
        return res;
    }
};

int main()
{
    vector<int> res={2,1,5,6,2,3};
    Solution s;
    cout<<s.largestRectangleArea(res)<<endl;
    return 0;
}