Given n>=0, create an array with the pattern {1, 1, 2, 1, 2, 3, ... 1, 2, 3 .. n} (spaces added to show the grouping). Note that the length of the array will be 1 + 2 + 3 ... + n, which is known to sum to exactly n*(n + 1)/2.
seriesUp(3) → [1, 1, 2, 1, 2, 3]
seriesUp(4) → [1, 1, 2, 1, 2, 3, 1, 2, 3, 4]
seriesUp(2) → [1, 1, 2]
Solution:
public int[] seriesUp(int n) {
int[] arr = new int[n*(n+1)/2];
int index = 0;
for(int i=0; i<n; i++){
for(int j=0; j<i+1;j++){
arr[index] = j+1;
index++;
}
}
return arr;
}