Leetcode stackAquene stack push and pop sequence

THE

problem

solution

Code

/*
思路: 这道题一开始我都看不懂。。。不知道找什么。看了一下分析才发现， 这道题的输入中间是可以穿插输出的。
那么整体的思路就是压入的时候判断是否等于输出的第一个了， 如果等于的话代表这个被弹出了，然后继续压入判断。

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*/

class Solution {
    
    
public:
   bool IsPopOrder(vector<int> pushV,vector<int> popV) {
    
    
       if(pushV.size()!= popV.size()) return false;
       stack<int> in ;
       
       int j = 0;
       for(int i = 0; i<pushV.size();i++){
    
    
           if(pushV[i]!=popV[j])in.push(pushV[i]);
           else {
    
    
               
               ++j;
               while((!in.empty())&&(in.top()==popV[j])){
    
    
                   in.pop();
                   ++j;
               }                    
           }
       }
       return in.empty();
       
       
   }
};

Summary and reflection

1. // Note while(!in.empty())&&(in.top()==popV[j]) and while((in.top()==popV[j])&&(!in.empty() ) The difference. The first judgment is empty and exit, but if you put j to the front, it will cause out of bounds. The logic is okay, but it is indeed out of bounds. You need to pay attention to the design in the future.