QUESTION:
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
nandkare in the range 1 <= k < n <= 10000.
EXPLANATION:
思路说一下：
从上面我们可以看出，如果有n个数，那么需要k个不同的数，正常应该是
k,k-1,k-2,k-3….
那么规律就来了，比如n=9，k=7的时候就是：
1，1+7，1+7-6；1+7-6+5
也就是总是等于前面一个数+或者- k的递减数列。
然后再进入第二个循环.
这样，我们
SOLUTION:
class Solution {
public int[] constructArray(int n, int k) {
int[] result = new int[n];
int index= 0;
int times = 0;
while (index<n){//当前数组还没有结束
int gap = 0;
result[index] = 1+times*(k+1);//将每次循环第一位数进行填充
index++;
times++;//进入下一个循环
if(result[index-1]+k<=n) gap = k;//如果gap还在n之内，那么就直接用k进行累减
else gap = n-result[index-1];//如果gap已经在n之外了，那么我们只能用n-result[index-1]作为gap了
boolean order = true;//标记当前需要做的是加还是减
while (gap > 0 && index <n){//每次k为gap的循环
result[index] = order? result[index-1]+gap:result[index-1]-gap;
gap--;
index++;
order = !order;
}
}
return result;
}
}