利用两点：

1. 0按位异或a=a

2. a按位异或a=0
   所以如下，复杂度是o(n)

def find(arr):
    tmp = 0
    for item in arr:
        tmp ^= item
    return tmp

if __name__ == '__main__':
    arr = [1,2,3,2,1,2,3,2,4]

时间复杂度O(n):

# coding: utf-8

from collections import defaultdict

def find_odds(arr):
    odds = defaultdict(int)
    for item in arr:
         odds[item] += 1
         if odds[item] > 1:
             del odds[item]
    return odds.keys()


arr = [5,1,2,3,2,1,2,3,2,4]
odds = find_odds(arr)