667. Beautiful Arrangement II

题目

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.

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.

分析

构造题，给两个数n和k，构造一个包含1~n的数组，让相邻两位的差的绝对值有k种。

     i++ j-- i++ j--  i++ i++ i++ ...
out: 1   9   2   8    3   4   5   6   7
dif:   8   7   6   5    1   1   1   1 

类似于上面的构造，最小值和最大值交叉的填入数组，到达k-1时让差绝对值为1（递减或递增）。这样做可以让前面的差绝对值横大于1,后面递增只添加1个新差绝对值；因为是最大最小交叉，中间的值还是有序的，也可以实现递增或递减
下面给出两种构造方法的代码，前者后面恒递增，后者后面可以递增或递减

代码

public int[] constructArray(int n, int k) {
    int[] res = new int[n];
    for (int i = 0, l = 1, r = n; l <= r; i++)
        res[i] = k > 1 ? (k-- % 2 != 0 ? l++ : r--) : l++;  //k奇偶时的变化体现在前面部分的构造
    return res;
    
    // int[] res = new int[n];
    // int i = 1,j = n, p = 0;
    // for(; p < k; ++p){
    //     if(p % 2 == 0){
    //         res[p] = i ++;
    //     }else{
    //         res[p] = j --;
    //     }
    // }
    // int q = p;
    // for(;p < n; ++p){
    //     res[p] = q % 2 == 1 ? i++ : j--;  //k奇偶时的变化体现在后面部分的构造
    // }
    // return res;
}