leetcode1 Two sum【基础题】【Python刷题】

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我是靠谱客的博主慈祥面包，最近开发中收集的这篇文章主要介绍leetcode1 Two sum【基础题】【Python刷题】，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

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].

十分愚蠢的错误QAQ

开始写成了

class Solution(object):
def twoSum(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: List[int]
"""
ans=[]
for i in range(len(nums)):
if(nums.count(target-nums[i])>0 ):
ans.append(i)
ans.append(nums.index(target-nums[i]))
break
return ans

发现这样会出现 [3,2,4] 6输出[0,0] 不满足use the same element twice.

后来改成了

class Solution(object):
def twoSum(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: List[int]
"""
ans=[]
for i in range(len(nums)):
if(nums.count(target-nums[i])>0 and target-nums[i] <>nums[i]):
ans.append(i)
ans.append(nums.index(target-nums[i]))
break
return ans

会出现[3,3]6什么都不输出QAQ

class Solution(object):
def twoSum(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: List[int]
"""
ans=[]
for i in range(len(nums)):
if(nums.count(target-nums[i])>0 and nums.index(target-nums[i]) <>i):
ans.append(i)
ans.append(nums.index(target-nums[i]))
break
return ans

这么简单的题写这么多遍才过QAQ