标签：Nging 解法 ack NPU -- vector from struct while
问题：
给定一个n，有数组1～n，
排列该数组，使得数组两两元素之间的差值有k种。
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.
解法：
如有以下数组
[1,2,3,4,5,6,7,8,9,10]
n=10,
k=9的情况，则有
i, j, i++, j--, i++, j--, i++, j--, i++, j--, [1, 10, 2, 9, 3, 8, 4, 7, 5, 6] k= 9， 8， 7， 6， 5， 4， 3， 2， 1
k=6的情况，则有
j, i, j--, i++, j--, i++, i++, i++, i++, i++ [10, 1, 9, 2, 8, 3, 4, 5, 6, 7] k= 9， 8， 7， 6， 5， 1， 1， 1， 1
代码参考：
1 class Solution { 2 public: 3 vector<int> constructArray(int n, int k) { 4 vector<int> res; 5 if(n<1||k<1)return res; 6 int i=1, j=n; 7 while(i<=j){ 8 if(k>1){ 9 res.push_back(k--%2?i++:j--); 10 }else{ 11 res.push_back(i++); 12 } 13 } 14 return res; 15 } 16 };
标签：Nging 解法 ack NPU -- vector from struct while
原文地址：https://www.cnblogs.com/habibah-chang/p/12731231.html
