leetCode —— Single Number

Given an array of integers, every element appears twice except for one. Find that single one.

Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

 

#-------------------------------------------------------------------------------

# Name:        module1

# Purpose:

#

# Author:      ScottGu<[email protected], [email protected]>

#

# Created:     13/11/2014

# Copyright:   (c) ScottGu<gu.kai.66> 2014

# Licence:     <your licence>

#-------------------------------------------------------------------------------



class Solution:

    # @param A, a list of integer

    # @return an integer

    def singleNumber(self, A):

        self.__init__()

        for num in A:

            if(self.dict.has_key(num)):

                self.dict[num]+=1

            else:

                self.dict[num]=1

        for p in self.dict.items():

            if(p[1]==1):

                return p[0]



    def __init__(self):

        self.dict={}

def main():

so=Solution()
arr=[1,1,2,2,3,3,4,4,5,6,6]
print arr
print so.singleNumber(arr)
arr=[1,2,2,3,3,4,4,5,6,6]
print arr
print so.singleNumber(arr)
arr=[1,1,2,3,3,4,4,5,6,6]
print arr
print so.singleNumber(arr)

arr=[1,1,2,2,3,3,4,4,5]
print arr
print so.singleNumber(arr)

arr=[1,0,1]
print arr
print so.singleNumber(arr)

arr=[1,0,0]
print arr
print so.singleNumber(arr)