【leetcode】1441. Build an Array With Stack Operations

题目如下：

Given an array target and an integer n. In each iteration, you will read a number from  list = {1,2,3..., n}.

Build the target array using the following operations:

• Push: Read a new element from the beginning list, and push it in the array.
• Pop: delete the last element of the array.
• If the target array is already built, stop reading more elements.

You are guaranteed that the target array is strictly increasing, only containing numbers between 1 to n inclusive.

Return the operations to build the target array.

You are guaranteed that the answer is unique. 

Example 1:

Input: target = [1,3], n = 3
Output: ["Push","Push","Pop","Push"]
Explanation: 
Read number 1 and automatically push in the array -> [1]
Read number 2 and automatically push in the array then Pop it -> [1]
Read number 3 and automatically push in 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"]
Explanation: You only need to read the first 2 numbers and stop.

Example 4:

Input: target = [2,3,4], n = 4
Output: ["Push","Pop","Push","Push","Push"]

Constraints:

• 1 <= target.length <= 100
• 1 <= target[i] <= 100
• 1 <= n <= 100
• target is strictly increasing.

解题思路：依次读取1~n之间的数字，如果数字等于target[0]，那么只要做Push操作，同时删掉target[0]；如果不相等，那么做Push和Pop操作。

代码如下：

class Solution(object):
    def buildArray(self, target, n):
        """
        :type target: List[int]
        :type n: int
        :rtype: List[str]
        """
        res = []
        for i in range(1,n+1):
            if len(target) == 0:
                break
            if i == target[0]:
                res.append("Push")
                target.pop(0)
            else:
                res.append("Push")
                res.append("Pop")

        return res