题目地址：https://leetcode.com/problems/two-sum/description/
Tag: Array
Difficulty: easy

题目描述

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:

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Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

解题思路

用一层for循环遍历nums数组中的元素x，令k = target-x，用in判断k是否在nums数组中，如果在，返回x和k

代码实现

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class Solution:
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        for i in range(len(nums)):
            k = target - nums[i]
            if k in nums and i != nums.index(k):
                index = nums.index(k)
                return [i,index]

第10行，i != nums.index(k)是为了防止nums=[3,3], target=6这样的情况，需要过滤掉k和nums[i]处在同一个位置。

第二种思路还有另一种实现方法，通过字典构建一个哈希表：

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class Solution:
    def twoSum(self, nums, target):
        dic = {}
        for i,num in enumerate(nums):
            if num in dic:
                return [dic[num],i]
            else:
                dic[target-num] = i

在Discuss区还看到一种清奇的实现方法，代码地址

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class Solution(object):
    def twoSum(self, nums, target):
        for i in range(len(nums)-1,0,-1):
            if target-nums[i] in nums[:i]:
                return [nums.index(target-nums[i]),i]