第一题：题目内容
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].
给定一个整数数组和一个目标值，找出数组中和为目标值的两个数。 你可以假设每个输入只对应一种答案，且同样的元素不能被重复利用。
示例: 给定 nums = [2, 7, 11, 15], target = 9
因为 nums[0] + nums[1] = 2 + 7 = 9 所以返回 [0, 1]
今天做了leetcode的第一道题，题意很简单，就是给定一组数和一个目标值，从这组数中找到两个数加起来等于目标值，输出目标值的下标。
刚刚见到这道题，没过脑，直接使用暴力遍历，两个循环嵌套，逐个进行相加运算，复杂度是O(n^2)
def twoSum( nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: List[int]
"""
l=list()
for i in range(len(nums)):
print('i=',i)
for j in range(i+,len(nums)):
print('j=',j)
if nums[i]+nums[j]==target:
l.append(i)
l.append(j)
return l
查找了别人写的之后感觉还是大神厉害的哈哈，自己提交之后AC了
def twoSum( nums, target):
sum=[]
for i in range(len(nums)):
one = nums[i]
second = target-nums[i]
if second in nums:
j = nums.index(second)
if i!=j:
sum.append(i)
sum.append(j)
return sum