已知一个整数列表，请编写一个函数，找出唯一一个出现奇数次数的那个数字。
提示：
- 总是只有一个整数出现奇数次。
题目难度：中等
题目来源：codewar
def find_it(nums:list):
pass
assert find_it([20,1,-1,2,-2,3,3,5,5,1,2,4,20,4,-1,-2,5]) == 5
assert find_it([1,1,2,-2,5,2,4,4,-1,-2,5]) == -1
assert find_it([10]) == 10
def find_it(nums: list):
result = []
for i in nums:
if i not in result:
result.append(i)
else:
result.remove(i)
if len(result) != 1:
return False
else:
return result[0]
assert find_it([20, 1, -1, 2, -2, 3, 3, 5, 5, 1, 2, 4, 20, 4, -1, -2, 5]) == 5
assert find_it([1, 1, 2, -2, 5, 2, 4, 4, -1, -2, 5]) == -1
assert find_it([10]) == 10
使用count函数 简单方便
def find_it(nums:list):
for i in nums:
res=nums.count(i)
if res%2==1:
return i
print(find_it([20, 1, -1, 2, -2, 3, 3, 5, 5, 1, 2, 4, 20, 4, -1, -2, 5]))
print(find_it([1,1,2,-2,5,2,4,4,-1,-2,5]))
print(find_it([10]))
class Solution:
@classmethod
方法一
def find_it(cls, nums: List[int]) -> int:
"""
1、定义字典，遍历列表，统计元素在列表中出现次数并保存到字典里.
2、遍历字典，判断次数为奇数的返回
:param nums
:return:int
"""
item_count = {}
for count in nums:
item_count[count] = nums.count(count)
for i, j in item_count.items():
if j % 2 == 0:
continue
return i
@classmethod
方法二
def find_its(cls, nums: List[int]) -> int:
"""
1、通过numpy将list转换成sclar list
2、统计出元素出现的次数放到字典
3、遍历字典，判断次数为奇数的返回
:param nums
:return: int
"""
arr = np.array(nums)
new_arr = np.unique(arr)
item_count = {}
for i in new_arr:
flag = (i == arr)
arr_new = arr[flag]
item_count[i] = arr_new.size
for key, value in item_count.items():
if value % 2 == 0:
continue
return key
assert_that(Solution.find_it(
[20, 1, -1, 2, -2, 3, 3, 5, 5, 1, 2, 4, 20, 4, -1, -2, 5]), equal_to(5))
assert_that(Solution.find_its(
[1, 1, 2, -2, 5, 2, 4, 4, -1, -2, 5]), equal_to(-1))
assert_that(Solution.find_its([10]), equal_to(10))
def find_it(nums:list):
for i in nums:
n = nums.count(i)
if n%2 != 0:
return i
assert find_it([20,1,-1,2,-2,3,3,5,5,1,2,4,20,4,-1,-2,5]) == 5
assert find_it([1,1,2,-2,5,2,4,4,-1,-2,5]) == -1
assert find_it([10]) == 10
def find_it(li):
""" 0716 循环奇数个的那个数字 """
# li = [20, 1, -1, 2, -2, 3, 3, 5, 5, 1, 2, 4, 20, 4, -1, -2, 5]
for i in li:
num = li.count(i)
if num%2 == 1:
return i
def find_it(nums: list):
res = {}
for i in nums:
if i in res.keys():
res[i] += 1
else:
res[i] = 1
for k, v in res.items():
if v % 2 != 0:
return k
assert find_it([20, 1, -1, 2, -2, 3, 3, 5, 5, 1, 2, 4, 20, 4, -1, -2, 5]) == 5
assert find_it([1, 1, 2, -2, 5, 2, 4, 4, -1, -2, 5]) == -1
assert find_it([10]) == 10
import pytest
@pytest.mark.parametrize("data,expect",[([20,1,-1,2,-2,3,3,5,5,1,2,4,20,4,-1,-2,5],5),
([1,1,2,-2,5,2,4,4,-1,-2,5],-1),
([10],10)])
def test_find_it(data,expect):
list1=list(set(data))
dict1={}
for i in list1:
dict1[i] = Counter(data)[i]
for i in dict1.values():
if i%2 !=0 :
assert {value:key for key,value in dict1.items()}[i] == expect
def find_it(nums:list):
return int([num for num in nums if nums.count(num) % 2 == 1 ][0])
assert find_it([20,1,-1,2,-2,3,3,5,5,1,2,4,20,4,-1,-2,5]) == 5
assert find_it([1,1,2,-2,5,2,4,4,-1,-2,5]) == -1
assert find_it([10]) == 10
def find_it(nums: list):
frequency_list = []
for n in nums:
if n in [i[0] for i in frequency_list if len(frequency_list) > 0]:
index = 0
for f in frequency_list:
if f[0] == n:
frequency_list[index][1] += 1
index += 1
else:
frequency_list.append([n, 1])
for f in frequency_list:
if f[1] % 2 == 1:
print(f[0])
return f[0]
assert find_it([20, 1, -1, 2, -2, 3, 3, 5, 5, 1, 2, 4, 20, 4, -1, -2, 5]) == 5
assert find_it([1, 1, 2, -2, 5, 2, 4, 4, -1, -2, 5]) == -1
assert find_it([10]) == 10
def find_it(nums: list):
dict = {}
for i in nums:
# 如果字典中有对应key值，则value值+1
if i in dict.keys():
dict[i] = int(dict[i]) + 1
else:
# 如果字典中没有对应key值，则插入，赋值1
dict[i] = 1
# 便利字典items，找出values值是奇数的key值
for i in dict.items():
if i[1] % 2 != 0:
return i[0]
assert find_it([20, 1, -1, 2, -2, 3, 3, 5, 5, 1, 2, 4, 20, 4, -1, -2, 5]) == 5
assert find_it([1, 1, 2, -2, 5, 2, 4, 4, -1, -2, 5]) == -1
assert find_it([10]) == 10
def find_it(nums: list):
for x in nums:
if nums.count(x) % 2 == 1:
return x
assert find_it([20, 1, -1, 2, -2, 3, 3, 5, 5, 1, 2, 4, 20, 4, -1, -2, 5]) == 5
assert find_it([1, 1, 2, -2, 5, 2, 4, 4, -1, -2, 5]) == -1
assert find_it([10]) == 10
def find_it(nums:list):
_dict = {}
for i in range(len(nums)):
if nums[i] in _dict: _dict[nums[i]].append(i)
else: _dict[nums[i]]=[i]
for key in _dict.keys():
if len(_dict[key])%2==1: return key
return None
第二种：比较拉的：
def find_it(nums:list):
s_list = sorted(nums)
start_index,end_index,start_w = -1,-1,None
for i in range(len(s_list)):
if start_w != s_list[i]:
if end_index != -1 and (end_index-start_index)%2==0:
return start_w
else:
start_w = s_list[i]
start_index = i
end_index = i
if (end_index-start_index)%2==0: return start_w
else: return None
def find_it(nums:list):
for num in set(nums):
if nums.count(num) % 2 == 1:
return num
assert find_it([20,1,-1,2,-2,3,3,5,5,1,2,4,20,4,-1,-2,5]) == 5
assert find_it([1,1,2,-2,5,2,4,4,-1,-2,5]) == -1
assert find_it([10]) == 10
def find_it(nums:list):
dict_nums={}
for i in range(len(nums)):
if nums[i] not in dict_nums.keys():
dict_nums[nums[i]]=1
else:
dict_nums[nums[i]]+=1
for j in dict_nums.items():
if j[1]%2 !=0:
return j[0]
assert find_it([20,1,-1,2,-2,3,3,5,5,1,2,4,20,4,-1,-2,5]) == 5
assert find_it([1,1,2,-2,5,2,4,4,-1,-2,5]) == -1
assert find_it([10]) == 10