Leetcode 1. Two Sum (Easy)
Description
Given an array of integers, return indices of the two numbers such that they add up to a specific target.
You may assume that each input would have exactly one solution, and you may not use the same element twice.
Example:
Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1].
Solution
每遍历一个nums[i],判断target - nums[i]是否在nums中。
这里用到dict(即tmp_num)来存储nums中每一个值及其对应index。
Notice
应该先判断target - nums[i]是否在tmp_num中,
再将nums[i]添加到tmp_num中,
否则若先将nums[i]添加到tmp_num,
则判断target - nums[i]时会将刚添加的nums[i]本身也算上。
错例:
input: [3, 2, 4] 6
output: [0, 0]
expected: [1, 2]
这里就是将刚添加的元素3算入了,应该先判断6 - 3是否在,再添加nums[0]。
Code
1 class Solution: 2 def twoSum(self, nums, target): 3 """ 4 :type nums: List[int] 5 :type target: int 6 :rtype: List[int] 7 """ 8 tmp_num = {} 9 for i in range(len(nums)): 10 if target - nums[i] in tmp_num: 11 return (tmp_num[target - nums[i]], i) 12 else: 13 tmp_num[nums[i]] = i; 14 return (-1, -1)
Beats: 46.65%
Runtime: 56ms