【每日一题1016】与众不同

兴趣爱好交流 每日一题
1

给定一个由正整数组成的列表，其中有且仅有一个与其他元素都不相同的数字，请编写一个函数，找出这个与众不同的数字。列表元素总个数是奇数个。

示例：
输入：[1, 1, 2]，输出，2
输入： [17, 17, 3, 17, 17, 17, 17]，输出：3

题目难度：简单
题目来源：codewars : Find the stray number

def find_stray(nums: list) -> int:
    pass

assert find_stray([1, 1, 1, 1, 1, 1, 2]) == 2
assert find_stray([2, 3, 2, 2, 2]) == 3
assert find_stray([3, 2, 2, 2, 2]) == 3
2
3
4 
def find_stray(numbers: list) -> int:
    for i in numbers:
        if numbers.count(i) == 1:
            return i


assert find_stray([1, 1, 1, 1, 1, 1, 2]) == 2
assert find_stray([2, 3, 2, 2, 2]) == 3
assert find_stray([3, 2, 2, 2, 2]) == 3
5 
def find_stray(nums: list) -> int:
    # 定义一个计数字典
    dict_count = dict()
    # 遍历列表中的每个数字，并统计出现次数
    for num in nums:
        # 获取该数字之前在计数字典中统计的次数
        # 如果没有被统计过则设置初始值为 0
        count = dict_count.get(num, 0)
        # 该数字统计次数 +1
        dict_count[num] = count + 1
    # 在计数字典中找出只出现一次的数字
    for k in dict_count:
        if dict_count[k] == 1:
            return k


assert find_stray([1, 1, 1, 1, 1, 1, 2]) == 2
assert find_stray([2, 3, 2, 2, 2]) == 3
assert find_stray([3, 2, 2, 2, 2]) == 3
6 
   public Integer findStray(List<Integer> list) {
        Integer res = list.get(0);
        for (int i = 1; i < list.size(); i++) {
            if (list.get(i) != res) {
                return list.get(i);
            }
        }
        return res;
    }

    @Test
    public void testFindStray() {
        assert 2 == findStray(Arrays.asList(1, 1, 2));
        assert 3 == findStray(Arrays.asList(17, 17, 3, 17, 17, 17, 17));
    }
7 
def find_stray(nums: list) -> int:
    for n in set(nums):
        if nums.count(n) == 1:
            return n


assert find_stray([1, 1, 1, 1, 1, 1, 2]) == 2
assert find_stray([2, 3, 2, 2, 2]) == 3
assert find_stray([3, 2, 2, 2, 2]) == 3
8 
def find_stray(nums: list) -> int:
    return [item for item in list(set(nums)) if nums.count(item) == 1][0]

assert find_stray([1, 1, 1, 1, 1, 1, 2]) == 2
assert find_stray([2, 3, 2, 2, 2]) == 3
assert find_stray([3, 2, 2, 2, 2]) == 3
10 
def find_stray(nums: list) -> int:
    pre = nums[0]
    next_ = nums[1]
    for i in nums[2:]:
        if i == pre and i == next_:
            pre = next_
            next_ = i
        elif i == next_:
            return pre
        elif i == pre:
            return next_
        else:
            return i


assert find_stray([1, 1, 1, 1, 1, 1, 2]) == 2
assert find_stray([2, 3, 2, 2, 2]) == 3
assert find_stray([3, 2, 2, 2, 2]) == 3
11

参考题解：

from functools import reduce

def find_stray(nums: list) -> int:
    return reduce(lambda x,y:x ^ y, nums)

assert find_stray([1, 1, 1, 1, 1, 1, 2]) == 2
assert find_stray([2, 3, 2, 2, 2]) == 3
assert find_stray([3, 2, 2, 2, 2]) == 3

思路：利用位运算中的异或来操作两个数字，如果相同数字则会得到0，任何数字和0进行异或运算会得到数字自身。因此，使用标准库中的reduce模块，将列表中的所有元素进行归约处理，当所有的相同元素消除得到0后，最终得到与众不同的唯一数字。

12 
def find_stray(nums: list) -> int:
    for i in nums:
        if nums.count(i)==1:
            return i

assert find_stray([1, 1, 1, 1, 1, 1, 2]) == 2
assert find_stray([2, 3, 2, 2, 2]) == 3
assert find_stray([3, 2, 2, 2, 2]) == 3
13 
def find_stray(nums: list) -> int:
    for x in nums:
        if nums.count(x)==1:
            return x


assert find_stray([1, 1, 1, 1, 1, 1, 2]) == 2
assert find_stray([2, 3, 2, 2, 2]) == 3
assert find_stray([3, 2, 2, 2, 2]) == 3