LeetCode 667. Beautiful Arrangement II

Posted 皇家大鹏鹏

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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 };

 


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