題目：

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

解釋：
這系列題目的第一題：
526. Beautiful Arrangement（python+cpp）
對於本題目，就是求1-n構成的陣列的一個序，這個序要滿足兩兩之間元素差的絕對值（相同的值算一種情況）構成的陣列大小為k。返回滿足條件的陣列序列。
嘗試著去找規律，發現是很簡單的一個規律題。首先我們確定的一點：元素差最多的個數是n-1個，這個n-1的構成也很容易發現，較大的數和較小的數交替形成的序列就滿足要求。例如，我們假設n= 6，k = 5，那麼這個序列就是 6 1 5 2 4 3 形成的k個元素差為： 5 4 3 2 1 （反向也是可以的 即 1 6 2 5 3 4）若k不等於n-1，我們只需要按上述規律形成滿足k-1的序列，剩餘序列按遞減序即可（剩餘的差值都為1）
假設n = 6，k = 4，我們得到的序列即是： 1 6 2 5 4 3。
所以要用雙指標做。
注意，上面的迴圈完成以後，只有k-2個，所以最後需要先往上跳一個變成k-1個，再遞減序列。上面的判斷一定是left所指的數結尾。
python程式碼：

class Solution(object):
    def constructArray(self, n, k):
        """
        :type n: int
        :type k: int
        :rtype: List[int]
        """
        left=1
        right=n
        result=[]
        while left<=right:
            if k>1:
                if k%2==0:
                    result.append(left)
                    left+=1
                else:
                    result.append(right)
                    right-=1
                k-=1
            else:
                result.append(right)
                right-=1
        return result           

c++程式碼：

class Solution {
public:
    vector<int> constructArray(int n, int k) {
        int left=1,right=n;
        vector<int> result;
        while (left<=right)
        {
            if (k>1)
                result.push_back((k--%2)==0?left++:right--);
            else
                result.push_back(right--);
        }
        return result;
    }
};

總結：
c++程式碼比python程式碼更加簡潔，恩。