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)