Description
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
- Only one valid answer exists.
Follow-up: Can you come up with an algorithm that is less than
O(n2)
time complexity?
Solution
Python3
class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
mp = {}
for i, x in enumerate(nums):
if target - x in mp:
return [mp[target - x], i]
mp[x] = i
Python
class Solution:
def twoSum(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: List[int]
"""
dict = {}
for i in range(len(nums)):
if target-nums[i] not in dict:
dict[nums[i]]=i
else:
return [dict[target-nums[i]],i]