946. Validate Stack Sequences -----M

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.

 

Constraints:

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

A:

Just simulate the process on generating the popped result.

每次要想pop一个值。 需要操作stack来完成。

class Solution {

public:

bool validateStackSequences(vector<int>& pushed, vector<int>& popped) {

stack<int> SS;

int start = 0;

for(auto val : popped){

while(start< popped.size() && (SS.empty() || SS.top() != val)){

SS.push(pushed[start++]);

}

if(SS.empty() || SS.top() != val){

return false;

}

SS.pop();

}

return true;

}

};

------------第二遍-------------------反而退步了----这个set是不需要的。并不需要检查head是否为下一个pop的，不是就继续存即可-----

class Solution {

public:

bool validateStackSequences(vector<int>& pushed, vector<int>& popped) {

stack<int> S;

unordered_set<int> mySet;

int idx = -1, n = pushed.size(); // index for pushed

// search each popped, and then see if we can make it pushed

for(auto val : popped){

while(idx < n && (mySet.find(val) == mySet.end())){

S.push(pushed[++idx]);

mySet.insert(pushed[idx]);

}

// now, we find it on the top of the stack

if(!S.empty() && S.top()==val){// no need to check empty actually

S.pop();

}else{

return false;

}

}

return true;

}

};