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---
title: leetcode-1-two-sum
tags: ["leetcode", "solution"]
date: 2019-11-24 21:06:37
---
> 虽然以前就就在Leetcode上面刷过题但是都没有坚持下去。这次准备重操旧业希望能起码做到一天一道题吧。
首先是A+B问题按理说是所有题目的开始。但是很遗憾的是我还是提交了三次才AC感觉自己弱爆了。废话不多说进入正题。
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```
1. Two Sum
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].
```
题意很简单就是给一个list然后给定一个target需要从list中找到两个数的和是target然后输出这两个数的下标。需要注意的是每个元素只能用一次另外输出的list中元素的顺序不影响结果。
我很快就完成了,但是第一版代码是错的,代码如:
```python
class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
m = dict()
for i in range(len(nums)):
if i not in m:
m[nums[i]] = i
if nums[i] in m and target-nums[i] in m:
return [m[nums[i]], m[target - nums[i]]]
return []
```
致命的问题在于如果test case是[3,2,1] 6那么输出是 [0,0],也就是说第一个元素用了两次,嗯,这么简单的边界都翻车。
然后很快又改好了,加了判断条件,代码如:
```python
class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
m = dict()
for i in range(len(nums)):
if nums[i] not in m:
m[nums[i]] = i
if nums[i] in m and target-nums[i] in m and m[nums[i]] != m[target - nums[i]]:
return [m[nums[i]], m[target - nums[i]]]
return []
```
然后又WA了。。。嗯仔细分析了一下发现test case是[3,3] 6的时候输出是空。嗯分支条件写错了。不应该判断两个数字都在字典里面因为如果有两个相同的元素的话只有一个会加入到字典中。为了解决这个问题直接换个写法就好
```python
class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
m = dict()
for i in range(len(nums)):
if nums[i] not in m:
m[nums[i]] = i
if target-nums[i] in m and i != m[target - nums[i]]:
return [i, m[target - nums[i]]]
```
总算AC了。我感觉每天一题的目标估计是要GG了。
稍微看了下题解和别人的代码,我发现这么写更加优雅:
```python
class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
m = dict()
for i, n in enumerate(nums):
if target-n in m:
return [m[target - n], i]
if n not in m:
m[n] = i
```
# Refer
[1. Two Sum](https://leetcode.com/problems/two-sum/)
[1. Two Sum](https://leetcode.com/articles/two-sum/)
# Appendix