leetcode 80 删除排序数组中的重复项 II

给定一个排序数组，你需要在原地删除重复出现的元素，使得每个元素最多出现两次，返回移除后数组的新长度。不要使用额外的数组空间，你必须在原地修改输入数组并在使用 O(1) 额外空间的条件下完成。

示例 1:

给定 nums = [1,1,1,2,2,3],
函数应返回新长度 length = 5, 并且原数组的前五个元素被修改为 1, 1, 2, 2, 3 。

示例 2:

给定 nums = [0,0,1,1,1,1,2,3,3],
函数应返回新长度 length = 7, 并且原数组的前五个元素被修改为 0, 0, 1, 1, 2, 3, 3 。

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• 双指针

  class Solution(object):
      def removeDuplicates(self, nums):
          """
          :type nums: List[int]
          :rtype: int
          """
          if len(nums) <= 2:
              return len(nums)
          prev = None
          cnt = 0
          total = 0
          p = 0
          for i in nums:
              if prev == None:
                  prev = i
                  cnt = 1
              elif prev == i:
                  cnt += 1
              else:
                  if cnt >= 2:
                      nums[p] = prev
                      nums[p+1] = prev
                      p += 2
                      prev = i
                      cnt = 1
                      total += 2
                  else:
                      nums[p] = prev
                      p += 1
                      prev = i
                      cnt = 1
                      total += 1
          if cnt >= 2:
              nums[p] = prev
              nums[p+1] = prev
              total += 2
          if cnt == 1:
              nums[p] = prev
              total += 1
          return total

• pattern

  class Solution:
      def removeDuplicates(self, nums， k):
          i = 0
          for n in nums:
              if i < k or n != nums[i-k]:
                  nums[i] = n
                  i += 1
          return i