给定一个由纯数字组成的列表，其中可能存在多个重复的数字，请编写一个函数，将相同数字中先出现的数字移除，只保留该重复数字的最后一个。返回去重后的列表。
示例：
输入：[3, 4, 4, 3, 6, 3]，输出：[4, 6, 3]。因为对于数字3，索引是0和3的会被移除；对于数字4，索引1的会被移除。
题目难度：简单
题目来源：CodeWars: Simple remove duplicates
def solution(nums: list) -> list:
# your code
assert solution([3,4,4,3,6,3]) == [4,6,3]
assert solution([1,2,1,2,1,2,3]) == [1,2,3]
assert solution([1,2,3,4]) == [1,2,3,4]
def solution(nums):
result_list=[]
for num in nums[::-1]:
if num not in result_list:
result_list.append(num)
return result_list[::-1]
def test(data:list):
new =
for i in data[::-1]:
if i not in new:
new.append(i)
new.reverse()
return new
def solution(nums: list) -> list:
nums.reverse()
num = []
for n in nums:
if n not in num:
num.append(n)
num.reverse()
return num
def solution(nums: list) -> list:
return sorted(list(set(nums[::-1])), key=nums[::-1].index)[::-1]
assert solution([3, 4, 4, 3, 6, 3]) == [4, 6, 3]
assert solution([1, 2, 1, 2, 1, 2, 3]) == [1, 2, 3]
assert solution([1, 2, 3, 4]) == [1, 2, 3, 4]
def solution(nums):
return list({}.fromkeys(nums[::-1]).keys())[::-1]
def solution(nums: list) -> list:
nums2 = [x for x in nums]
for i in range(len(nums) - 1):
if nums[i] in nums[i + 1:]:
nums2.remove(nums[i])
return nums2
assert solution([3, 4, 4, 3, 6, 3]) == [4, 6, 3]
assert solution([1, 2, 1, 2, 1, 2, 3]) == [1, 2, 3]
assert solution([1, 2, 3, 4]) == [1, 2, 3, 4]
def solution(nums: list) -> list:
l = []
for i in range(len(nums)-1,-1,-1):
if nums[i] not in l:
l.insert(0,nums[i])
return l
def solution15(nums:list):
result=[]
for i in nums[::-1]:
if i not in result:
result.insert(0,i)
return result
assert solution15([3,4,4,3,6,3]) == [4,6,3]
assert solution15([1,2,1,2,1,2,3]) == [1,2,3]
assert solution15([1,2,3,4]) == [1,2,3,4]
def Removal():
list = [3, 4, 4, 3, 6, 3]
listRemoval=
for i in list:
if i not in listRemoval:
listRemoval.append(i)
print(listRemoval)
Removal()
def solution(nums: list) -> list:
aft_nums = []
for idx, num in enumerate(nums):
if num not in nums[idx + 1:]:
aft_nums.append(num)
return aft_nums
assert solution([3, 4, 4, 3, 6, 3]) == [4, 6, 3]
assert solution([1, 2, 1, 2, 1, 2, 3]) == [1, 2, 3]
assert solution([1, 2, 3, 4]) == [1, 2, 3, 4]
def solution(nums: list) -> list:
print(nums[::-1])
n = 0
m = len(nums) - 1
while n < m:
for i in range(0, len(nums)-1):
if nums[i] in nums[i+1:]:
nums.pop(i)
break
n += 1
return nums
def solution(nums: list) -> list:
# 方法一
nums.reverse()
new = []
for i in nums:
if i not in new:
new.append(i)
new.reverse()
return new
# 方法二
# 利用 fromkeys 创建新字典，seq 为反转后的 nums
# 这样会得到自动去重后的k-v，只取 key，再做反转
return list(dict.fromkeys(nums[::-1]).keys())[::-1]
# 方法三
# 利用 sorted 排序规则，按照反转后的列表下标排序
return sorted(list(set(nums[::-1])), key=nums[::-1].index)[::-1]
方法 2 和方法 3 是看到评论区的解法，学到了，
。
def solution(nums: list) -> list:
result=[]
for i in nums:
if i in result:
result.remove(i)
result.append(i)
return result
assert solution([3,4,4,3,6,3]) == [4,6,3]
assert solution([1,2,1,2,1,2,3]) == [1,2,3]
assert solution([1,2,3,4]) == [1,2,3,4]
``def solution(nums: list) -> list:
# your code
for i in set(nums):
while nums.count(i) > 1:
nums.remove(i)
return numsdef solution(nums: list) -> list:
return [x for i,x in enumerate(nums) if nums[i+1:].count(x)==0]
def Removal():
list = [3, 4, 4, 3, 6, 3]
list2 =
for i in list:
if i not in list2:
list2.append(i)
elif i in list2:
list2.remove(i)
list2.append(i)
print(list2)
Removal()
def solution(nums: list) -> list:
list1 = list[::-1]
list2 =[]
for i in list1:
if i not in list2:
list2.append(i)
return list2[::-1]def solution(nums: list) -> list:
added = set()
return [i for i in nums[::-1] if not (i in added or added.add(i))][::-1]
assert solution([3,4,4,3,6,3]) == [4,6,3]
assert solution([1,2,1,2,1,2,3]) == [1,2,3]
assert solution([1,2,3,4]) == [1,2,3,4]
def solution(nums: list) -> list:
# your code
list_num = []
for i in nums[::-1]:
if i not in list_num:
list_num.append(i)
return list_num[::-1]
assert solution([3, 4, 4, 3, 6, 3]) == [4, 6, 3]
assert solution([1, 2, 1, 2, 1, 2, 3]) == [1, 2, 3]
assert solution([1, 2, 3, 4]) == [1, 2, 3, 4]