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].
思路：
将列表的索引和值保存到map中，判断差值是否也存在的列表中，可以达到O(n)复杂度。
Python实现：
class Solution(object):
def twoSum(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: List[int]
"""
map = {}
for i, val in enumerate(nums): # 遍历列表的下标和索引，也可以使用 for i in range(len(nums))
gap = target - val
if gap in map: # 如果键存在于map中
return [map[gap], i]
else:
map[val] = i # 按照{值:索引}的形式保存到map中
return None
a = [2,3,5,7,11]
b = Solution()
print(b.twoSum(a[:], 16)) # [2, 4]
