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]
Output: 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?
class Solution {
List<int> twoSum(List<int> numbers, int target) {
Map<int, int> indexMap = {};
for (int i = 0; i < numbers.length; i++) {
int requiredNum = target - numbers[i];
if (indexMap.containsKey(requiredNum)) {
return [indexMap[requiredNum]!, i];
}
indexMap[numbers[i]] = i;
}
return [-1, -1];
}
}