1. Two Sum 經典老題目

 1. Two Sum

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
• Only one valid answer exists.

 

Follow-up: Can you come up with an algorithm that is less than O(n2) time complexity?

這題真的是經典，可以說是leetcode的開宗始祖

直覺是O(n^2) 算法，但如果用一個字典先store 整個array

就可以變成O(n)算法 每次就去查找index即可，真令人懷念

class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        
        
        
        d=dict()
        for i in range(len(nums)):
            d[nums[i]]=i
        
        for j in range(len(nums)):
            if target-nums[j] in d and d[target-nums[j]]!=j:
                return [j,d[target-nums[j]]]