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 <= 109Follow-up: Can you come up with an algorithm that is less than
O(n2) time complexity?1class Solution {
2 public int[] twoSum(int[] nums, int target) {
3 for(int i=0;i<nums.length-1;i++){
4 int complement=target-nums[i];
5 for(int j=i+1;j<nums.length;j++){
6 if (complement==nums[j]){
7 return new int []{i,j};
8 }
9 }
10 }
11 return new int[]{};
12 }
13}
141class Solution:
2 def twoSum(self, nums: List[int], target: int) -> List[int]:
3 numMap = {}
4 n = len(nums)
5
6 for i in range(n):
7 complement = target - nums[i]
8 if complement in numMap:
9 return [numMap[complement], i]
10 numMap[nums[i]] = i
11
12 return [] # No solution found