leetcode 667. Beautiful Arrangement II

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.

构造。很明显，如果序列是1,2,3....n会有1个不同。这里我们只要改变后面两个数字的顺序就会得到{2,1}两个不同的数。如果交换3个数比如1,2,3,4 得到 1,4,2,3 会有{3,2,1}3个不同。其实，我们只要k和i构造k-i的差值就行了，比如n = 4,k = 3 那么我们可以构造差值 3 2 1就行了。所以序列是 1 4,2,3。比如n=5,k=3我们构造差值3,2,1

所以序列是   1,4,2,3,5

O(n^2)代码。

class Solution {
public:
    vector<int> constructArray(int n, int k) {
        vector<int> a(n);
        for (int i = 0; i < n; ++i) {
            a[i] = i + 1;
        }
        if (k == 1) return a;
        for (int i = 0; i < k - 1; ++i) {
            reverse(a.begin() + i + 1, a.end());
        }
        return a;
    }
};

O(n)代码

class Solution {
public:
    vector<int> constructArray(int n, int k) {
        vector<int> ans;
        int l=1,r=k+1;
        while(l<=r)
        {
            ans.push_back(l);
            l++;
            if(l<r) ans.push_back(r--);
        }
        for(int i=k+2; i<=n; i++)
            ans.push_back(i);
        
        return ans;
    }
};