查找列表中唯一的唯一值[重复]

Question

我正在参加一项 Codewars 挑战，以寻找列表中的唯一值，但我无法以我的技能水平找到完成此过程的好方法。

此解决方案适用于测试，但尝试超时（12000 毫秒）。

def find_uniq(arr):
    for element in arr:
        if arr.count(element) == 1:
            return element

练习链接： https://www.codewars.com/kata/585d7d5adb20cf33cb000235/train/python

Answer 1

您的算法是 O(n^2)，其中 n 是数组元素的数量。您应该考虑使用dict 或collections.Counter。

# Not tested
def findUnique(arr):
  from collections import Counter
  counter = Counter(arr)
  for value, count in counter.items():
    if count == 1:
      return value

Answer 2

一个简单的方法是跟踪每个项目的计数。

from collections import Counter

def find_uniq(arr):
    # Count each item.
    # I suggest you try to write your own Counter as an exercise.
    # You will need to loop over the entire array.
    # Store your results in a dict.
    counts = Counter(arr)

    # Now, loop over the resulting counts and check which item has a count of 1.
    for value, count in counts.items():
        if count == 1:
            return value

    raise ValueError("Could not find unique item!")

Answer 3

基于相邻元素排序和检查的解决方案

def find_uniq(arr: list):
    arr.sort()
    n = len(arr) - 1
    for i, x in enumerate(arr):
        if (i == n or x != arr[i + 1]) and (
                i == 0 or x != arr[i - 1]):  # if there are no identical neighboring elements
            return x
    return None

print(find_uniq([10, 10, 10, 1, 1, 1, 1, 1, 100, 2, 2, 2, 2, 3, 3, 3, 3, 7, 7, 7, 7, 8, 8, 8, 8]))

打印：

100