LeetCode 667. Beautiful Arrangement II

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, … , a_n], then the list [|a₁ - a₂|, |a₂ - a₃|, |a₃ - a₄|, … , |a_n-1 - a_n|] 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.

題目分析：給定n和k，要求輸出一個包含1-n的序列，序列滿足規律：相鄰數差的絕對值得種類恰好是k個。
輸出序列必然存在多個（證明：假設序列q符合題意，那麼reverse(q)必然也成立）。現在假設這k個差表示的是1到k，並且相鄰差的絕對值也是按照這個順序。構造的符合題意得數列如下：

 

 k=1:   1  2  3  4  5  6  7  

 k=2:   2  1  3  4  5  6  7 

 k=3:   2  3  1  4  5  6  7 

 k=4:   3  2  4  1  5  6  7 

 k=5:   3  4  2  5  1  6  7 

 …….. 

 

規律：
1、p[0]=k/2+1;
2、deta=1到k，正負號相間
3、若k是奇數，則deta先正後負，否則先負後正。
7、以上規律持續到deta=k，即第（k+1)個位置，之後的數據順序遞增。

代碼如下：

class Solution {
    public int[] constructArray(int n, int k) {
        int[]result=new int[n];
        int st=(k/2)+1;
        boolean flag=k%2==0?false:true;
        result[0]=st;
        for(int i=1;i<=k;i++){
            if(flag)
                result[i]=result[i-1]+i;
            else
                result[i]=result[i-1]-i;
            flag=!flag;
        }
        for(int i=k+1;i<n;i++)
            result[i]=i+1;
        return result;
    }
}