题目：原题链接（中等）
标签：数组、双指针
解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
Ans 1 (Python) | O ( N ) | O ( 1 ) | 36ms (90.85%) |
Ans 2 (Python) | |||
Ans 3 (Python) |
解法一：
class Solution:
_SMALL = -10001
def removeDuplicates(self, nums: List[int]) -> int:
i1 = 0
now_val, now_num = self._SMALL, 0
for i2 in range(len(nums)):
if nums[i2] != now_val:
nums[i1] = nums[i2]
now_val, now_num = nums[i2], 1
i1 += 1
elif now_num < 2:
nums[i1] = nums[i2]
now_num += 1
i1 += 1
else:
now_num += 1
nums[i1:] = []
return len(nums)
