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.
#-*- coding:utf-8 -*-
class Solution:
# @param {integer[]} nums
# @return {integer}
def removeDuplicates(self, nums):
l = len(nums)
if l == 0:
return 0
index = 0
i = 1
nums[index] = nums[0]
while i<l:
if nums[index] != nums[i]:
index += 1
nums[index] = nums[i]
i += 1
return index+1
if __name__=="__main__":
s = Solution()
print s.removeDuplicates(x)
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。该方法通过双指针技巧实现,其中一个指针用于跟踪当前唯一元素的位置,另一个指针遍历整个数组。当遇到新的不同元素时,将其复制到唯一元素指针位置,并递增该指针。最终返回唯一元素的数量。
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