双指针秒杀数组去重
双指针算法原理
双指针算法通过维护两个指针（通常称为快指针和慢指针）在数组中协同工作。快指针用于遍历数组，慢指针用于标记有效元素的边界。该算法能在O(n)时间复杂度内完成操作，空间复杂度为O(1)。
问题分析（LeetCode 26题）
给定一个升序排列的数组，要求原地删除重复元素，使每个元素只出现一次，并返回新数组的长度。操作需保证空间复杂度为O(1)，即不借助额外数据结构。
算法实现步骤
初始化慢指针slow为0，快指针fast从1开始遍历数组。
当nums[fast]不等于nums[slow]时，将slow右移一位，并将nums[fast]的值赋给nums[slow]。
遍历结束后，slow + 1即为新数组长度。
def removeDuplicates(nums):
if not nums:
return 0
slow = 0
for fast in range(1, len(nums)):
if nums[fast] != nums[slow]:
slow += 1
nums[slow] = nums[fast]
return slow + 1
边界条件处理
空数组直接返回0。
数组长度为1时无需处理，直接返回1。
复杂度分析
时间复杂度：O(n)，仅需一次线性遍历。
空间复杂度：O(1)，仅使用常量级额外空间。
变体与扩展
若允许元素最多出现k次，可将条件改为nums[fast] != nums[slow - k + 1]。例如k=2时：
def removeDuplicatesK(nums, k=2):
slow = k - 1
for fast in range(k, len(nums)):
if nums[fast] != nums[slow - k + 1]:
slow += 1
nums[slow] = nums[fast]
return slow + 1
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