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.

判断栈的合法性

class Solution {
public:
bool validateStackSequences(vector<int>& pushed, vector<int>& popped) {
if(pushed.size()==&&pushed.size()==){
return true;
}
if(pushed.size()!=popped.size()){
return false;
}
stack<int>s;
int index = ,outdex = ;
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())//如果入栈序列已经走完,出栈序列和栈顶元素一一比较
{
while (!s.empty() && popped[outdex]==s.top())
{
s.pop();
outdex++;
} //如果和栈中比较完,两个序列都走完了,即表明顺序合法
if (index == pushed.size() && outdex == popped.size())
{
return true;
}
else
{
return false;
}
}
}
}
};
 

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