首页 > 代码库 > leetCode —— Single Number
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<gu.kai.66@gmail.com, 150316990@qq.com>## 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)
leetCode —— Single Number
声明：以上内容来自用户投稿及互联网公开渠道收集整理发布，本网站不拥有所有权，未作人工编辑处理，也不承担相关法律责任，若内容有误或涉及侵权可进行投诉： 投诉/举报 工作人员会在5个工作日内联系你，一经查实，本站将立刻删除涉嫌侵权内容。