思想: 1.判断列表nums的长度是否小于等于1,若是则返回列表nums的长度,反之跳转2 2.定义一个变量i=0,代表不重复元素的下标。定义for循环,j从1依次变化到len(nums) 3.判断nums[i]是否与nums[j]相等。若否,则i加1,nums[j]赋值给nums[i]。跳转3,直到j==len(nums)-1结束
class Solution:
def removeDuplicates(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
if len(nums) <= 1:
return len(nums)
else:
i = 0
for j in range(1, len(nums)):
if nums[i] != nums[j]:
i += 1
nums[i] = nums[j]
return i+1