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

原文地址：https://www.cnblogs.com/shiyublog/p/9424488.html