Leetcode simple problem building arrays with stack operation

subject

Give you an array of targets target And an integer n. Every iteration , Need from list = {1,2,3…, n} Read a number in sequence .
Use the following operations to build the target array target ：
Push： from list Read a new element in , And push it into the array .
Pop： Delete the last element in the array .
If the target array is built , Stop reading more elements .
The title data ensures that the target array is strictly incremented , And only contains 1 To n Number between .
Please return the operation sequence used to build the target array .
The question data ensures that the answer is unique .
Example 1：
Input ：target = [1,3], n = 3
Output ：[“Push”,“Push”,“Pop”,“Push”]
explain ：
Read 1 And automatically push the array -> [1]
Read 2 And automatically push the array , Then delete it -> [1]
Read 3 And automatically push the array -> [1,3]
Example 2：
Input ：target = [1,2,3], n = 3
Output ：[“Push”,“Push”,“Push”]
Example 3：
Input ：target = [1,2], n = 4
Output ：[“Push”,“Push”]
explain ： Just before reading 2 A number can stop .
Example 4：
Input ：target = [2,3,4], n = 4
Output ：[“Push”,“Pop”,“Push”,“Push”,“Push”]
Tips ：
1 <= target.length <= 100
1 <= target[i] <= 100
1 <= n <= 100
target It's strictly incremental
source ： Power button （LeetCode）

Their thinking

   Since the given target It's strictly incremental , Then we just need to traverse the generated sequence from front to back , When the encountered value is not equal to the current target Elements , Then press it directly into the stack and then pop it up , If it is equal to , Just push into the stack . Or transform the sequence into a queue , Just go out and compare .

class Solution:
    def buildArray(self, target: List[int], n: int) -> List[str]:
        operating=[]
        i=1
        j=0
        while i<n+1 and j<len(target):
            if i==target[j]:
                operating.append('Push')
                j+=1
                i+=1
            else:
                operating.extend(['Push','Pop'])
                i+=1
        return operating

class Solution:
    def buildArray(self, target: List[int], n: int) -> List[str]:
        operating=[]
        Q=deque(range(1,n+1))
        i=0
        while Q and i<len(target):
            temp=Q.popleft()
            if temp==target[i]:
                operating.append('Push')
                i+=1
            else:
                operating.extend(['Push','Pop'])
        return operating