LeetCode 667. Beautiful Arrangement II

Posted 2020-10-15 皇家大鹏鹏

 

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:

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

 

 这道题非常巧妙，考虑的是数字排列的 问题。如果暴力搜索，复杂度是O(n*n！)，必然会超时，因此需要另辟蹊径，题目要求1-n之间的n个数，按照某个顺序形成了一个排列，相邻两个数做差取绝对值，形成一个新的向量，新的向量中只能有k个不同的数字，求n个数的组合。我们不难发现，k最大取值为n-1

我们其实可以反过来考虑，构造1,2，...，k+1序列，恰好可以使得差值有1,2，...，k个不同数字

1，k+1，2，k，.....

这时如果数字还没用完，那么从k+2开始遍历到n，相邻数字之间的差值均为1，不产生新的差值，满足题目要求

代码如下：

 1 class Solution {
 2 public:
 3     vector<int> constructArray(int n, int k) {
 4         int l = 1, r = k+1;
 5         vector<int> ans;
 6         while (l <= r) {
 7             ans.push_back(l++);
 8             if (l <= r) ans.push_back(r--);
 9         }
10         for (int i = k+2; i <= n; i++)
11             ans.push_back(i);
12         return ans;
13     }
14 };