667 Beautiful Arrangement II

667. Beautiful Arrangement II

1. Question

Given two integersnandk, you need to construct a list which containsndifferent positive integers ranging from1tonand 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 exactlykdistinct 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. Thenandkare in the range 1 <= k < n <= 10^4.

2. Implementation

思路:

(1) 当K等于1时，显然直接将1-n升序放在数组即可

(2) 当K大于1时，我们可以发现K最大的可能值是n - 1，比如当n = 5时，数组为

1， 2， 3， 4，5

得到最多4个差值得数组为

1， 5， 2， 4，3

差: 4, 3, 2, 1

我们发现规律是，将1 - n的数交错放在数组里，比如先放1，再放n, 然后放2，之后放n - 1，以此类推

class Solution {
    public int[] constructArray(int n, int k) {
        if (n == 0 || k <= 0 || k >= n) {
            return new int[0];
        }

        int[] res = new int[n];
        int start = 1, end = n;

        for (int i = 0; start <= end; i++) {
            // 当 K 等于 1时，将1-n的数字直接升序放在数组即可
            if (k== 1) {
                res[i] = start;
                ++start;
            }
            // 否则将start和end交错放在数组里
            else {
                if (k % 2 != 0) {
                    res[i] = start;
                    ++start;
                }
                else {
                    res[i] = end;
                    --end;
                }
                --k;
            }
        }
        return res;
    }
}

3. Time & Space Complexity

时间复杂度O(n), 空间复杂度O(n)