LeetCode-667：Beautiful Arrangement II (数组的完美安排) -- medium

Question

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 = 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:

• The n and k are in the range 1 <= k < n <= 104.

问题解析：

给定整数n和k，构造一个包含1-n的n个整数的数组[a1, a2, a3, … , an]，使得[|a1 - a2|, |a2 - a3|, |a3 - a4|, … , |an-1 - an|] 新数组含有k个不同的数。

Answer

Solution 1：

二分法。

• 由题我们可知：1..n最多可以构造出n-1个不同的差，比如 1..9为：[1 9 2 8 3 7 4 6 5]
• 其相邻元素差的绝对值为：diff: 8 7 6 5 4 3 2 1
• 观察构造的数据，其是大小交替的。那么这样的话，我们只要先构造出前k个，后面按照顺序来产生1就可以了。
• 在后面顺序添加的时候注意，需要添加的是增序还是逆序。
• 如：以1..7为例：
• k=6：1 7 2 6 3 5 | 4
• k=5：1 7 2 6 3 | 4 5
• k=4：1 7 2 6 | 5 4 3
• k=3：1 7 2 | 3 4 5 6
• k=2：1 7 | 6 5 4 3 2
• k=1:1 | 2 3 4 5 6 7

class Solution {    public int[] constructArray(int n, int k) {        int[] ans = new int[n];        int l = 1, r = n;        int i = 0;        for (; i < k; i++){            if (i % 2 == 0) ans[i] = l++;            else ans[i] = r--;        }        if(i % 2 == 1){            for (int j = l; j <= r; j++) ans[i++] = j;        }else{            for (int j = r; j >= l; j--) ans[i++] = j;        }        return ans;    }}

• 时间复杂度：O(n)，空间复杂度：O(n)