Pop up and push in sequence of stack ＜ difficulty coefficient ＞

Title Description ： Enter two sequences of integers , The first sequence represents the push order of the stack , Please determine if the second sequence is likely to be the pop-up sequence of the stack . Suppose that all the Numbers pushed are not equal . For example, the sequence 1,2,3,4,5 Is the push order of a stack , Sequence 4,5,3,2,1 Is a popup sequence corresponding to the stack sequence , but 4,5,3,2,1 It cannot be a popup sequence of the push sequence .

Tips ：

• 0 <= pushV.length == popV.length <= 1000
• -1000 <= pushV[i] <= 1000
• pushV All numbers are different

Example 1：

Input ：
[1,2,3,4,5],[4,5,3,2,1]
 
Return value ：
true
 
explain :
Can pass
push(1) => push(2) => push(3) => push(4) => pop() => push(5) => pop() => pop() => pop() => pop()
In this order, we get [4,5,3,2,1] This sequence , return true.

Example 2：

Input ：
[1,2,3,4,5],[4,3,5,1,2]

Return value ：false

explain ：
Because it is [1,2,3,4,5] Press in sequence of ,[4,3,5,1,2] Pop up order , requirement 4,3,5 Must be in 1,2 Front press in , And 1,2 Can't pop up , But in this order ,1 Can't in 2 Pop up before , So it can't form , return false.

🧷 platform ：Visual studio 2017 && windows

The core idea ： We have come across multiple-choice questions before this question . The essence of this problem is to simulate the characteristics of the stack “Last In First Out”.

Here, a stack is defined to simulate , It doesn't matter ,pushi First in stack , And then ++, The order of getting out of the stack must be after entering the stack , So you need to judge every time you enter the stack pushi and popi Whether it is equal or not , Equal and equal ( And cycle out ), Otherwise, enter , Both of them can go to the end , It means it's a match .

nowcoder The original title is

class Solution {public:    bool IsPopOrder(vector<int> pushV,vector<int> popV) {        stack<int> st;        size_t pushi = 0, popi = 0;        // When the pressing sequence ends, the result must be         while(pushi < pushV.size())        {             // Enter a data first , then ++            st.push(pushV[pushi++]);            // Loop out the stack             while(!st.empty() && st.top() == popV[popi])            {                ++popi;                st.pop();            }        }        //st It's empty , Description match         return st.empty();        // ditto         //return popi == popV.size();    }};