LeetCode 1. Two Sum (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:

Given nums = [2, 7, 11, 15], target = 9,Because nums[0] + nums[1] = 2 + 7 = 9,return [0, 1].

题目大意：给出一个数组，找出值相加等于target的两个数字并输出其下标。

思路：暴力法：对于每个数组元素，判断剩余元素与其相加是否为target，复杂度O(n^2)
哈希法：设当前元素为x。遍历数组，如果target - x存在于哈希表，则输出其下标，复杂度O(n)，即遍历数组的复杂度。因为我对c++的map不熟，就用python的dict来做了。
python代码（哈希）：

class Solution(object):    def twoSum(self, nums, target):        """        :type nums: List[int]        :type target: int        :rtype: List[int]        """        dict = {}        for i in range(len(nums)):            if dict.get(target - nums[i], None) == None:                dict[nums[i]] = i            else:                return (dict[target - nums[i]], i)