003.数组中重复的数字

# -*- coding:utf-8 -*-
from collections import Counter
class Solution(object):
    def findRepeatNum(self,input_list):
        if not  input_list:
            return
        c = Counter(input_list)
        result = set()
        for index,num in enumerate(input_list):
            if c[num] >= 2:
                result.add(num)
        return result
    def find2(self,input_list):
        if not input_list:
            return
        #step 1 : sorted
        input_list.sort()
        result = set()
        #
        for index in  range(len(input_list)-1):
            if input_list[index] == input_list[index+1]:
                result.add(input_list[index])
        return result
        #step 2:iter
    def find3(self,input_list):
        ##check input
        if not input_list:
            return
        for num in input_list:
            if num <0 or num > (len(input_list)-1):
                return
        res = set()
        for index in range(0,len(input_list)):
            #如果index代表的元素，如果和index不等的话则处理
            while input_list[index] != index:
                # #如果input_list[index] == input_list[input_list[index]] 添加到res
                if input_list[index] == input_list[input_list[index]] :
                    res.add(input_list[index])
                    break #跳出循环
                #交换
                else:
                    tmp = input_list[index]
                    input_list[index] = input_list[tmp]
                    input_list[tmp] = tmp

        return res
        ##after check
s = Solution()
print s.find3([3,1,2,2,2,5,3])
#find1:space:O(n) time:O(1)
#find2:space:O(1) time:O(nlog(n))
#find3:space:O(1) time:O(n)