正文
Python 循环列表删除元素的注意事项
提示：扫一扫查出行【扫一扫了解最新限行尾号】
复制提示
错误示范：
class Solution:
def removeElement(self, nums, val: int) -> int:
for i, num in enumerate(nums):
print('i=', i, ', num=', num, ', nums=', nums)
if num == val:
nums.remove(val)
return len(nums) s = Solution()
s.removeElement([2, 0,1,2,2,3,0,4,2], 2)
# i= 0 , num= 2 , nums= [2, 0, 1, 2, 2, 3, 0, 4, 2]
# i= 1 , num= 1 , nums= [0, 1, 2, 2, 3, 0, 4, 2]
# i= 2 , num= 2 , nums= [0, 1, 2, 2, 3, 0, 4, 2]
# i= 3 , num= 3 , nums= [0, 1, 2, 3, 0, 4, 2]
# i= 4 , num= 0 , nums= [0, 1, 2, 3, 0, 4, 2]
# i= 5 , num= 4 , nums= [0, 1, 2, 3, 0, 4, 2]
# i= 6 , num= 2 , nums= [0, 1, 2, 3, 0, 4, 2]
解决方式：
① 使用尾递归方式
class Solution:
def removeElement(self, nums, val: int) -> int:
for i, num in enumerate(nums[::-1]):
print('i=', i, ', num=', num, ', nums=', nums)
if num == val:
nums.remove(val)
return len(nums) s = Solution()
s.removeElement([2, 0,1,2,2,3,0,4,2], 2)
# i= 0 , num= 2 , nums= [2, 0, 1, 2, 2, 3, 0, 4, 2]
# i= 1 , num= 4 , nums= [0, 1, 2, 2, 3, 0, 4, 2]
# i= 2 , num= 0 , nums= [0, 1, 2, 2, 3, 0, 4, 2]
# i= 3 , num= 3 , nums= [0, 1, 2, 2, 3, 0, 4, 2]
# i= 4 , num= 2 , nums= [0, 1, 2, 2, 3, 0, 4, 2]
# i= 5 , num= 2 , nums= [0, 1, 2, 3, 0, 4, 2]
# i= 6 , num= 1 , nums= [0, 1, 3, 0, 4, 2]
# i= 7 , num= 0 , nums= [0, 1, 3, 0, 4, 2]
# i= 8 , num= 2 , nums= [0, 1, 3, 0, 4, 2]
② 使用 while 循环的方式
class Solution:
def removeElement(self, nums, val: int) -> int:
i = 0
while i < len(nums):
print('i=', i, ', num=', nums[i], ', nums=', nums)
if nums[i] == val:
nums.pop(i)
else:
i += 1
return len(nums) s = Solution()
s.removeElement([2, 0,1,2,2,3,0,4,2], 2)
# i= 0 , num= 2 , nums= [2, 0, 1, 2, 2, 3, 0, 4, 2]
# i= 0 , num= 0 , nums= [0, 1, 2, 2, 3, 0, 4, 2]
# i= 1 , num= 1 , nums= [0, 1, 2, 2, 3, 0, 4, 2]
# i= 2 , num= 2 , nums= [0, 1, 2, 2, 3, 0, 4, 2]
# i= 2 , num= 2 , nums= [0, 1, 2, 3, 0, 4, 2]
# i= 2 , num= 3 , nums= [0, 1, 3, 0, 4, 2]
# i= 3 , num= 0 , nums= [0, 1, 3, 0, 4, 2]
# i= 4 , num= 4 , nums= [0, 1, 3, 0, 4, 2]
# i= 5 , num= 2 , nums= [0, 1, 3, 0, 4, 2]
③ 对整个序列使用切片来创建一个临时副本
class Solution:
def removeElement(self, nums, val: int) -> int:
for i, num in enumerate(nums[:]):
print('i=', i, ', num=', num, ', nums=', nums)
if num == val:
nums.remove(val)
return len(nums) s = Solution()
s.removeElement([2, 0,1,2,2,3,0,4,2], 2)
# i= 0 , num= 2 , nums= [2, 0, 1, 2, 2, 3, 0, 4, 2]
# i= 1 , num= 0 , nums= [0, 1, 2, 2, 3, 0, 4, 2]
# i= 2 , num= 1 , nums= [0, 1, 2, 2, 3, 0, 4, 2]
# i= 3 , num= 2 , nums= [0, 1, 2, 2, 3, 0, 4, 2]
# i= 4 , num= 2 , nums= [0, 1, 2, 3, 0, 4, 2]
# i= 5 , num= 3 , nums= [0, 1, 3, 0, 4, 2]
# i= 6 , num= 0 , nums= [0, 1, 3, 0, 4, 2]
# i= 7 , num= 4 , nums= [0, 1, 3, 0, 4, 2]
# i= 8 , num= 2 , nums= [0, 1, 3, 0, 4, 2]
