Easy
Given an array of integers nums
and an integer target
, return indices of the two numbers such that they add up to target
.
You may assume that each input would have exactly one solution, and you may not use the same element twice.
You can return the answer in any order.
Example 1:
Input: nums = [2,7,11,15], target = 9
Output: [0,1]
Explanation: Because nums[0] + nums[1] == 9, we return [0, 1].
Example 2:
Input: nums = [3,2,4], target = 6
Output: [1,2]
Example 3:
Input: nums = [3,3], target = 6
Output: [0,1]
Constraints:
2 <= nums.length <= 104
-109 <= nums[i] <= 109
-109 <= target <= 109
Follow-up: Can you come up with an algorithm that is less than O(n2)
time complexity?
object Solution {
def twoSum(nums: Array[Int], target: Int): Array[Int] = {
val lookup = scala.collection.mutable.Map[Int, Int]()
for (idx <- nums.indices) {
if (lookup.contains(target - nums(idx))) {
return Array(lookup(target - nums(idx)), idx)
} else {
lookup.addOne(nums(idx) -> idx)
}
}
Array(-1, -1)
}
}