Given two integers n and k, you need to construct a list which contains n different 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.

這道題非常巧妙，考慮的是數字排列的 問題。如果暴力搜索，復雜度是O(n*n！)，必然會超時，因此需要另辟蹊徑，題目要求1-n之間的n個數，按照某個順序形成了一個排列，相鄰兩個數做差取絕對值，形成一個新的向量，新的向量中只能有k個不同的數字，求n個數的組合。我們不難發現，k最大取值為n-1

我們其實可以反過來考慮，構造1,2，...，k+1序列，恰好可以使得差值有1,2，...，k個不同數字

1，k+1，2，k，.....

這時如果數字還沒用完，那麽從k+2開始遍歷到n，相鄰數字之間的差值均為1，不產生新的差值，滿足題目要求

代碼如下：

 1 class Solution {
 2 public:
 3     vector<int> constructArray(int n, int k) {
 4         int l = 1, r = k+1;
 5         vector<int> ans;
 6         while (l <= r) {
 7             ans.push_back(l++);
 8             if (l <= r) ans.push_back(r--);
 9         }
10         for (int i = k+2; i <= n; i++)
11             ans.push_back(i);
12         return ans;
13     }
14 };

LeetCode 667. Beautiful Arrangement II