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, ... , 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.

 * 注意到：1..n最多可以弄出n-1个不同的差，比如1..9就是
 * 192837465
 * diff: 8 7 6 5 4 3 2 1
 假设要n=9 k=6  那么产生1   9   2   8   3   7  6   5   4   先产生k个数，也就是k-1个差值，然后剩下的n-k个数就需要按照差值为1进行产出，但是要特别提示的是，最后这几个数的产生是依赖于第k个数的位置，因为这这个数组中有两串，一个顺序，一个逆序，所以这里需要判断一下，再进行循环。

public static int[] constructArray(int n, int k) {        int []res=new int[n];    int l=1,r=n,i=0;    for(;i<k;i++){//先产生k个数    if(i%2==0) res[i]=l++;    else   res[i]=r--;    }    //将剩余的n-k个数放到数组中    if(i%2==1){//因为此时缺位为奇数位，但是前一位为偶数位，要达到差值为1，此时要自然序数递增产生余下的数    for(int j=k;j<n;j++) res[j]=l++;    }else {//因为此时缺位为偶数数位，但是前一位为奇数位，要达到差值为1，此时要r--for(int j=k;j<n;j++) res[j]=r--;}            return res;            }