由于题目说了有且只有唯一解,可以考虑两遍扫描求解:第一遍扫描原数组,将所有的数重新存放到一个dict中,该dict以原数组中的值为键,原数组中的下标为值;第二遍扫描原数组,对于每个数nums[i]查看target-nums[i]是否在dict中,若在则可得到结果。 
当然,上面两遍扫描是不必要的,一遍即可,详见代码。

class Solution(object):
def twoSum(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: List[int]
"""
keys = {}
for i in xrange(len(nums)):
if target - nums[i] in keys:
return [keys[target - nums[i]], i]
if nums[i] not in keys:
keys[nums[i]] = i

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