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。
第一关：
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]
class Solution:
def twoSum(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: List[int]
"""
for k,v in enumerate(nums):
if target - v in nums[k+1:]:
return [k,nums.index(target - v) ]
第二关：
示例:
[3,3]
class Solution:
def twoSum(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: List[int]
"""
for k,v in enumerate(nums):
if target - v in nums[k+1:]:
return [k,nums[k+1:].index(target - v) + k + 1]
第三关：
多个组合1440 ms
示例:
nums = [1,2,3,4,5,6,7,8,9]
target = 10
class Solution:
def twoSum(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: List[int]
"""
lst = []
for k,v in enumerate(nums):
if target - v in nums[k+1:]:
lst.append([k,nums[k+1:].index(target - v) + k + 1])
if len(lst) == 1:
lst = lst[0]
return lst
运行最快代码52ms
class Solution:
def twoSum(self, nums, target):
dic = {}
lst = []
for i in range(len(nums)):
if target-nums[i] in dic:
lst.append([dic[target-nums[i]], i])
dic[nums[i]]=i
if len(lst) == 1:
lst = lst[0]
return lst
