Description
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 <= 10^4.
分析
题目的大致意思是:给你一个整数的返回n,和一个k,然后把这n个数排成一圈,然后相邻见的差值的种类数为k,然后输出这样一个序列。
- 比如说当n=8,我们有数组:
1, 2, 3, 4, 5, 6, 7, 8
当我们这样有序排列的话,相邻两数的差的绝对值为1。我们想差的绝对值最大能为多少,应该是把1和8放到一起,为7。那么为了尽可能的产生不同的差的绝对值,我们在8后面需要放一个小数字,比如2,这样会产生差的绝对值6,同理,后面再跟一个大数,比如7,产生差的绝对值5,以此类推,我们得到下列数组:
1, 8, 2, 7, 3, 6, 4, 5
其差的绝对值为:7,6,5,4,3,2,1
共有7种,所以我们知道k最大为n-1,所以这样的排列一定会存在。我们的策略是,先按照这种最小最大数相邻的方法排列,没排一个,k自减1,当k减到1的时候,后面的排列方法只要按照生序的方法排列,就不会产生不同的差的绝对值,这种算法的时间复杂度是O(n),属于比较高效的那种。我们使用两个指针,初始时分别指向1和n,然后分别从i和j取数加入结果res,每取一个数字k自减1,直到k减到1的时候,开始按升序取后面的数字
代码
class Solution {
public:
vector<int> constructArray(int n, int k) {
vector<int> res;
int i=1;
int j=n;
while(i<=j){
if(k>1){
if(k%2==1){
res.push_back(i);
i++;
}else{
res.push_back(j);
j--;
}
k--;
}else{
res.push_back(i++);
}
}
return res;
}
};