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
