LC 667. Beautiful Arrangement II

最新推荐文章于 2025-08-15 18:53:39 发布

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最新推荐文章于 2025-08-15 18:53:39 发布

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文章标签： c/c++ runtime

原文链接：http://www.cnblogs.com/ethanhong/p/10190519.html

本文介绍了一个算法问题，即如何构造一个包含n个不同正整数的列表，这些整数范围在1到n之间，且其相邻元素差值的绝对值列表中恰好包含k个不同的整数。文章提供了代码实现，并通过两个例子展示了输入与输出的关系。

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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 kdistinct 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.

Runtime: 28 ms, faster than 39.80% of C++ online submissions for Beautiful Arrangement II.

构造题

最小，k=1是顺序或者倒序的排列。

最大，k=n-1是依此从两头挑数的排列。

根据k来判断要排多少个头尾数。

#include <vector>
#include <iostream>
using namespace std;
class Solution {
public:
  vector<int> constructArray(int n, int k) {
    int tmp = k/2;
    vector<int> ret;
    int idx = 1;
    while(tmp){
      ret.push_back(idx);
      ret.push_back(n+1 - idx);
      idx++;
      tmp--;
    }
    if(k % 2 == 1){
      for(int i = idx; i<= n+1-idx; i++) ret.push_back(i);
    }else for(int i=n+1-idx; i>=idx; i--) ret.push_back(i);
    return ret;
  }
};