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?
func twoSum(nums []int, target int) []int {
indexMap := make(map[int]int)
for i, num := range nums {
requiredNum := target - num
if j, ok := indexMap[requiredNum]; ok {
return []int{j, i}
}
indexMap[num] = i
}
return []int{-1, -1}
}