PAT 1041 Be Unique python解法

1041 Be Unique （20 分）
Being unique is so important to people on Mars that even their lottery is designed in a unique way. The rule of winning is simple: one bets on a number chosen from [1,10^​4]. The first one who bets on a unique number wins. For example, if there are 7 people betting on { 5 31 5 88 67 88 17 }, then the second one who bets on 31 wins.

Input Specification:
Each input file contains one test case. Each case contains a line which begins with a positive integer N (≤10⁵​​ ) and then followed by N bets. The numbers are separated by a space.

Output Specification:
For each test case, print the winning number in a line. If there is no winner, print None instead.

Sample Input 1:
7 5 31 5 88 67 88 17
Sample Output 1:
31
Sample Input 2:
5 888 666 666 888 888
Sample Output 2:
None

解题思路：题意是找第一个与其他数不相同的数，输入的一行数据中，第一个数n是一共有几个数，后面跟着n个数。
第一个思路：建立两个列表，unique储存不相同的数，same存储去掉重复值之后的数，对输入l进行循环判断，如果l中元素在unique中，则将unique中该数删除，如果l中元素不在same中，则将该元素添加到same和unique中，循环结束后，如果unique中有元素，则输出第一个元素，如果为空，则输出None。此方法在2个测试点中超时。

#2个测试点运行超时
l = list(map(int,input().split()))
#l = [5, 888, 666, 666, 888, 888]
#l = [7, 5, 31, 5, 88, 67, 88, 17]
d = {}
for i in l[1:]:
    if i not in d.keys():
        d[i] = 1
    else:
        d[i] += 1
out = 0
for i in l[1:]:
    if i in d.keys() and d[i] == 1:
        out = i
        break
if out:
    print(out)
else:
    print('None')

第二个思路：建立一个字典d，对输入l进行循环判断，如果l中元素不在字典d中，则添加到字典d中，键为该元素，值为1，如果在字典d中，则将对应值加1。再对列表进行循环判断，找到第一个在字典d中的元素后跳出循环并将该元素保存到out中。out的初始值为0，因为题目所给的数据在[1,10^​4]，0不在其中。最后判断out是否为0，如果为0，则输出None，否则输出out。

l = list(map(int,input().split()))
#l = [5, 888, 666, 666, 888, 888]
#l = [7, 5, 31, 5, 88, 67, 88, 17]
d = {}
for i in l[1:]:
    if i not in d.keys():
        d[i] = 1
    else:
        d[i] += 1
out = 0
for i in l[1:]:
    if i in d.keys() and d[i] == 1:
        out = i
        break
if out:
    print(out)
else:
    print('None')