Given two sequences pushed and popped with distinct values, return true if and only if this could have been the result of a sequence of push and pop operations on an initially empty stack.

Example 1:

Input: pushed = [1,2,3,4,5], popped = [4,5,3,2,1]
Output: true
Explanation: We might do the following sequence:
push(1), push(2), push(3), push(4), pop() -> 4,
push(5), pop() -> 5, pop() -> 3, pop() -> 2, pop() -> 1
 

Example 2:

Input: pushed = [1,2,3,4,5], popped = [4,3,5,1,2]
Output: false
Explanation: 1 cannot be popped before 2.

Note:

1. 0 <= pushed.length == popped.length <= 1000
2. 0 <= pushed[i], popped[i] < 1000
3. pushed
    is a permutation of popped.
4. pushed and popped have distinct values.

判斷棧的合法性

class Solution {
public:
    bool validateStackSequences(vector<int>& pushed, vector<int>& popped) {
        if(pushed.size()==0
 
&&pushed.size()==0){
            return true;
        }
        if(pushed.size()!=popped.size()){
            return false;
        }
        stack<int>s;
        int index = 0,outdex = 0;
        while(index<pushed.size()&&outdex<=popped.size()){
            if(pushed[index]==popped[outdex]){
                index++;
                outdex++;
            }else if(!s.empty()&&popped[outdex]==s.top()){
                while(!s.empty()&&popped[outdex]==s.top()){
                    s.pop();
                    outdex++;
                }
            }else{
                s.push(pushed[index]);
                index++;
            }
            if (index >= popped.size())//如果入棧序列已經走完，出棧序列和棧頂元素一一比較
            {
                while (!s.empty() && popped[outdex]==s.top())
                {
                    s.pop();
                    outdex++;
                }

                //如果和棧中比較完，兩個序列都走完了，即表明順序合法
                if (index == pushed.size() && outdex == popped.size())
                {
                    return true;
                }
                else
                {
                    return false;
                }
            }
        }
    }
};