题目如下：

Given two integers n and k, you need to construct a list which contains ndifferent positive integers ranging from 1 to n and obeys the following requirement: 

Suppose this list is [a1, a2, a3, ... , an], then the list [|a1 - a2|, |a2 - a3|, |a3 - a4|, ... , |an-1 - an|] has exactly k distinct integers.

If there are multiple answers, print any of them.

Example 1:

Input: n = 3, k = 1Output: [1, 2, 3]Explanation: The [1, 2, 3] has three different positive integers ranging from 1 to 3, and the [1, 1] has exactly 1 distinct integer: 1.

 

Example 2:

Input: n = 3, k = 2Output: [1, 3, 2]Explanation: The [1, 3, 2] has three different positive integers ranging from 1 to 3, and the [2, 1] has exactly 2 distinct integers: 1 and 2.

 

Note:

1. The n and k are in the range 1 <= k < n <= 104.

解题思路：感觉自己对“给定一个条件，输出对应排列”这一类型的题目比较没有思路。本题的解法我也是想了很久才想出来。以[1,2,3,4,5,6]为例，当前k是等于1的；假设要k=2,只需要把6插入到头部[6,1,2,3,4,5]即可。而如果要k=3,那么把6插入到1的后面，变成[1,6,2,3,4,5]。接下来就是递推了，记插入6的位置为inx，要使得k=4，只要在k=2的基础上把最后一个元素插入到inx+2的位置；同理，如果是k=5，就在k=3的基础上操作。

代码如下：

class Solution(object):    def constructArray(self, n, k):        """        :type n: int        :type k: int        :rtype: List[int]        """        res = [i for i in range(1,n+1)]        inx = 0        if k % 2 == 1:            inx = 1        else:            res.insert(0, res.pop(-1))            k -= 1            inx += 2        while k > 1:            res.insert(inx,res.pop(-1))            inx += 2            k -= 2        return res