Interleave two lists
Two queues for two registers are interleaved so every other customer comes from each.
- Take one from the first list, then one from the second, and so on.
- When one list runs out, append whatever remains of the other.
alternatingMerge(first: list<int>, second: list<int>) → list<int>
Go needs a compiler and Drill does not host one yet, so this page is the reference rather than an exercise: the problem, worked examples, and the solution in full. To type it out, the same problem runs in Python.
Where you start
func alternatingMerge(first []int, second []int) []int {
}Worked examples
| Call | Result |
|---|---|
alternatingMerge([]int{1, 2, 3}, []int{9}) | []int{1, 9, 2, 3} |
alternatingMerge([]int{1}, []int{4, 5}) | []int{1, 4, 5} |
alternatingMerge([]int{1, 2}, []int{3, 4}) | []int{1, 3, 2, 4} |
alternatingMerge([]int{}, []int{1, 2}) | []int{1, 2} |
Hint
Loop up to the longer length and take each element that still exists.
Reference solution in Go
func alternatingMerge(first []int, second []int) []int {
result := []int{}
n := len(first)
if len(second) > n {
n = len(second)
}
for i := 0; i < n; i++ {
if i < len(first) {
result = append(result, first[i])
}
if i < len(second) {
result = append(result, second[i])
}
}
return result
}