题目：
'''
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.'''1. 最开始的写法：
class Solution:
def twoSum(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: List[int]
"""
if len(nums) <1 :
return false
for i in range(len(nums)):
temp=target-nums[i]
if temp in nums.index
while j<len(nums):
sum=nums[i]+nums[j]
if sum==target:
return i,j
else:
j+=1
大部分例子都能通过，有一个出现了如下图所示的提示：
2. 利用index改写为:
class Solution:
def twoSum(self, nums, target):
if len(nums) <1 :
return false
for i in range(len(nums)):
temp=target-nums[i]
if temp in nums:
if i!=nums.index(temp):
return i,nums.index(temp)3. 利用enumerate()来改写：
class Solution:
def twoSum(self, num, target):
d = dict()
for index, number in enumerate(num):
try:
return ((d[target-number]+1, index+1))
except:
d[number] = index看到有加了try的，感觉这样很好，借鉴一下。
enumerate(sequence, [start=0])- sequence,序列或迭代对象等
- start 下标起始位置
- 返回值：返回枚举对象，
>>> sq=['one','two','three']
>>> for index,element in enumerate(sq):
print(index,sq[index])
0 one
1 two
2 three
>>> 4.
class Solution:
# @return a tuple, (index1, index2)
# 8:42
def twoSum(self, num, target):
map = {}
for i in range(len(num)):
if num[i] not in map:
map[target - num[i]] = i + 1
else:
return map[num[i]], i + 1
return -1, -1