1,n,2,n-1,3,n-2,4... ==> Diff: n-1, n-2, n-3, n-4, n-5...

By following this pattern, k numbers will have k-1 distinct difference values;

and all the rest numbers should have |ai - a_i-1| = 1;

In total, we will have k-1+1 = k distinct values.

```
class Solution {
public int[] constructArray(int n, int k) {
if(k>=n) return null;
int[] arr = new int[n];
int i = 0, small = 1, large = n;
while(i<k){
arr[i++] = small++;
if(i<k) arr[i++] = large--;
}
if(k%2 == 0){ // k==2 ==> 1, 6, 5,4,3,2
while(i<arr.length) arr[i++] = large--;
} else { // k==3 ==> 1,6,2,3,4,5
while(i<arr.length) arr[i++] = small++;
}
return arr;
}
}
```