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"]
class Solution { public List<String> buildArray(int[] target, int n) { List<String> res = new ArrayList(); List<Integer> list = new ArrayList(); for(int i: target) list.add(i); for(int i = 1; i <= n; i++){ if(i > target[target.length - 1]) return res; if(list.indexOf(i) < 0){ res.add("Push"); res.add("Pop"); } else { res.add("Push"); } } return res; } }
多种方法,比如把target存到arraylist里面,然后遍历从1-n,当n比target最大的还大,就没有必要继续了,返回即可
内循环中,如果target里面妹有n,就push再pop,否则push