Problem:
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 being1and2respectively. It doesn’t matter what you leave beyond the new length.
Idea:
1. Kindly note that extra space cannot be allocated. Therefore, we can only remove items in the list. And the lenth of list will change due to removal.
2. After removing items, we should pay attention that the next item will probably be skipped if we donnot pay attention to the index. One solution is that we can traverse the list in a reverse order.
Solution:
class Solution(object):
def removeDuplicates(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
lennums = len(nums)
if lennums == 0:
return 0
else:
preitem = nums[0]
i = 1
while i < lennums:
if preitem == nums[i]:
del(nums[i])
lennums -= 1
else:
preitem = nums[i]
i += 1
return lennums
ref:
http://blog.youkuaiyun.com/lanchunhui/article/details/50984630
http://www.cnblogs.com/bananaplan/p/remove-listitem-while-iterating.html

本文介绍了一个不使用额外空间,仅通过原地修改来去除已排序数组中重复元素的方法,并提供了Python实现代码。该方法保留每个元素只出现一次,并返回新长度。

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