目前想根据GitHub上的LeetCode的整理，打算每天练几道题，增加自己的编程能力，当成课外习题去学习，希望能够有所进步和提升。

原题连接： https://leetcode.com/problems/two-sum

内容描述：

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].

解题方案
1、暴力解法，两轮遍历

class Solution(object):
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype List[int]
        """
        for i in range(len(nums)):
            for j in nums[i+1:]:
                if nums[i] + nums[j] == target:
                    return [i, j]

上面的思路1太慢了，我们可以牺牲空间换取时间
2、时间复杂度: O(N)，空间复杂度: O(N)

• 建立字典 lookup 存放第一个数字，并存放该数字的 index
• 判断 lookup 中是否存在： target - 当前数字， 则表面当前值和 lookup中的值加和为 target.
• 如果存在，则返回： target - 当前数字的index和当前值的index

class Solution(object):
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype List[int]
        """
        lookup = {}
        for i, num in enumerate(nums):
            if target - num in lookup.keys():
                return [lookup[target-num], i]
            else:
                lookup[num] = i