Leetcode Two Sum

#title : Two Sum
#Given an array of integers, find two numbers such
#that they add up to a specific target number.
#The function two Sum should return indices of the
#two numbers such that they add up to the target,
#where index1 must be less than index2. Please note
#that your returned answers (bothindex1 and index2)
#are not zero-based.
#You may assume that each input would have exactly
#one solution.
#Input: numbers={2, 7, 11, 15}, target=9
#Output: index1=1, index2=2

# First method O(n^2)
#easy but Time Limit Exceeded
# class Solution:
# 	# @return a tuple,(index1, index2)
# 	def twoSum(self, num, target):
# 		for i in range(len(num)):
# 			for j in range(i+1,len(num)):
# 				if (num[i] + num[j] == target):
# 					return (i+1,j+1)

#Second method O(n) HashTable
#HashTable store the number's index
class Solution:
# @return a tuple,(index1, index2)
def twoSum(self, num, target):
num_hashtable = dict()
result_tuple = tuple()
mul_hash = dict()
for i in range(len(num)):
#the key don't repeat
if num[i] in num_hashtable:
mul_hash[num[i]] = i
else:
#the key do repeat
num_hashtable[num[i]] = i
print mul_hash
print num_hashtable
for i in range(len(num)):
gap = target - num[i]
#two cases
#first : don't have same number
if gap >= 0 and gap != num[i]:
if num_hashtable.has_key(gap):
result_tuple = (i+1,num_hashtable[gap]+1)
return result_tuple
#second : have same number
elif gap >= 0 and gap == num[i]:
if gap in mul_hash:
result_tuple = (i+1,mul_hash[gap]+1)
return result_tuple

if __name__ == '__main__':
twoSum = Solution()
num = [3,2,5,7,3]
#	num = [0,0,3,2,8,9,7,8,9,7]
#	num = 0
target = 6
print twoSum.twoSum(num, target)