題目如下：

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 = 1
Output: [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 = 2
Output: [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.

解題思路：感覺自己對“給定一個條件，輸出對應排列”這一型別的題目比較沒有思路。本題的解法我也是想了很久才想出來。以[1,2,3,4,5,6]為例，當前k是等於1的；假設要k=2,只需要把6插入到頭部[6,1,2,3,4,5]即可。而如果要k=3,那麼把6插入到1的後面，變成[1,6,2,3,4,5]。接下來就是遞推了，記插入6的位置為inx，要使得k=4，只要在k=2的基礎上把最後一個元素插入到inx+2的位置；同理，如果是k=5，就在k=3的基礎上操作。

程式碼如下：

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