分類：Array
考察知識點：Array(數(shù)組遍歷)
最優(yōu)解時間復雜度：O(n)
26. Remove Duplicates from Sorted Array
Given a sorted array nums, 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 by modifying the input array in-place with O(1) extra memory.
Example1:
Given 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 returned length.
Example2:
Given nums = [0,0,1,1,1,2,2,3,3,4],
Your function should return length = 5, with the first five elements of nums being modified to 0, 1, 2, 3, and 4 respectively.
It doesn't matter what values are set beyond the returned length.
代碼：
我的解法:
class Solution:
def removeDuplicates(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
start,end=1,len(nums)
while (end>start):
if nums[start]==nums[start-1]:
nums.pop(start)
end-=1
else:
start+=1
return end
官方解法:
class Solution:
def removeDuplicates(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
if len(nums)==0:
return 0
else:
i=0
for j in range(1,len(nums)):
if nums[j]!=nums[i]:
i+=1
nums[i]=nums[j]
return i+1
討論：
1.這個題目修改了好多遍击胜，叫我們用inplace的方法缠黍，反正就是挺簡單的
2.直接改比pop效果好歪架，看來pop還是耗了一定的時間的
