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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:
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Input: n = 3, k = 1
2
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Output: [1, 2, 3]
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Explanation:
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The [1, 2, 3] has three different positive integers ranging from 1 to 3, and the [1, 1] has exactly 1 distinct integer: 1.
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Example 2:
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Input: n = 3, k = 2
2
3
Output: [1, 3, 2]
4
5
Explanation:
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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.
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Note:
  1. 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
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class Solution {
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public int[] constructArray(int n, int k) {
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if (n == 0 || k <= 0 || k >= n) {
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return new int[0];
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}
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int[] res = new int[n];
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int start = 1, end = n;
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for (int i = 0; start <= end; i++) {
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// 当 K 等于 1时,将1-n的数字直接升序放在数组即可
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if (k== 1) {
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res[i] = start;
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++start;
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}
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// 否则将start和end交错放在数组里
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else {
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if (k % 2 != 0) {
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res[i] = start;
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++start;
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}
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else {
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res[i] = end;
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--end;
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}
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--k;
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}
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}
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return res;
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}
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}
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3. Time & Space Complexity

时间复杂度O(n), 空间复杂度O(n)