Stack push-in, pop-up sequence

Title Description

Enter a sequence of two integers, and the first sequence represents the stacking order of the stack, judging whether the second sequence is the pop-up order for the stack. Assume that all the numbers that are pressed into the stack are not equal. For example, the sequence 1,2,3,4,5 is the indentation order of a stack, and the sequence 4,5,3,2,1 is a pop-up sequence corresponding to the stack sequence, but 4,3,5,1,2 is not likely to be the pop-up sequence of the stack sequence.

Solution:

Operation of the simulation stack

Create a stack variable that represents the stack;

2. If the stack is empty, or stack.top () ==popv[i],stack.pop

4. Otherwise, Stack.push (PUSHV "Inpos])

6. 
   

7.  bool IsPopOrder(vector<int> pushV, vector<int> popV) {
9. int size1 = Pushv.size ();
11. int size2 = Popv.size ();
12.     if (size1 != size2||size1==0)
13.         return false;
14.     stack<int> st;
15.     int inpos = 0;
16.     for (int i = 0; i<size2;) {
18. if (!st.empty () && st.top () = = Popv[i]) {
20. st.pop ();
22. i++;
24. }
26. else {
28. if (Inpos < size1)
30. st.push (pushv[inpos++]);
32.             else
33.                 return false;
35. }
37. }
39. return inpos = = Size1&&st.empty ();
41. }

Stack push-in, pop-up sequence