Given a sorted array, remove the duplicates in place such that each element appear only once and return the new length.
Do not allocate extra space for another array, you must do this in place with constant memory.
For example,
Given input array nums = [1,1,2],
Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively.
It doesn't matter what you leave beyond the new length.
class Solution(object):
def removeDuplicates(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
if len(nums) <= 1:
return len(nums)
else:
left = 0
right = left + 1
length =len(nums) - 1
while right < length:
if nums[left] == nums[right]:
temp = right
nums.pop(right)
right = temp
length = len(nums)-1
else:
left += 1
right += 1
if nums[left] == nums[right]:
nums.pop(right)
return len(nums)
本文介绍了一种在不使用额外空间的情况下从已排序数组中去除重复元素的方法,并返回处理后的数组长度。该方法通过双指针技术实现,左指针用于记录不同元素的位置,右指针用于遍历数组。当发现新元素时,将其移动到左指针位置并更新左指针。
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