--- title: leetcode-1-two-sum tags: ["leetcode", "solution"] date: 2019-11-24 21:06:37 --- > 嗯,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。 ``` 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,嗯,直接换个写法就好,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 target-nums[i] in m and i != m[target - nums[i]]: return [i, m[target - nums[i]]] ``` 嗯,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