Remove Duplicates from Sorted Array

Topic:

Given a sorted array, remove the duplicates in place such, all element appear only once and return the new L Ength.

Do the allocate extra space for another array, and you must does this on place with constant memory.

For example,
Given input array nums = [1,1,2] ,

Your function should return length = 2 , with the first of the elements of nums being and 1 2 Respectivel Y. It doesn ' t matter what are you leave beyond the new length.

Code:

There is no difficulty in understanding the topic, that is, the number of the order, remove the repetition, there are still a few.

However, the topic requires that memory cannot be increased, that is, you cannot create a new data structure, but you can delete an existing list:)

thus

#-*-Utf-8-*-
#Remove Duplicates from Sorted Array
Class solution (Object):
def removeduplicates (self, nums):
Total_length = Len (nums)
For Index,value in Enumerate (nums):
If index = = Total_length-1:break

#遇到后一个元素和该元素一样, delete the element until it is not the same
While Nums[index]==nums[index+1]:
List.remove (Nums, Nums[index])
#index = index+1
Total_length = total_length-1
If index = = Total_length-1:break
#print Nums
Return total_length

If __name__== "__main__":
NUMS=[-49,-49,-48,-48,-47,-47,-47,-46,-46,-46,-46,-44,-44,-44,-42,-42,-41,-41,-40,-40,-39,-39,-39,-38,-37,-37, -36,-36,-35,-35,-35,-34,-34,-32,-32,-31,-30,-29,-28,-28,-27,-27,-27,-26,-26,-25,-25,-25,-24,-23,-22,-21,-21,- 21,-20,-20,-20,-20,-20,-18,-18,-17,-17,-16,-16,-15,-15,-15,-14,-13,-13,-11,-10,-10,-9,-9,-9,-9,-9,-9,-7,-6,-6, -6,-5,-5,-5,-4,-3,-3,-1,-1,-1,-1,- 1,1,3,3,5,5,5,5,6,6,7,8,8,8,8,9,9,10,10,13,14,14,14,15,15,15,16,17,18,18,19,19,20,20,20,21,21,21,22,23,23,24,25,25,25,25 , 26,26,26,26,26,27,27,27,27,27,27,27,28,29,29,30,30,30,30,31,31,32,32,34,35,37,38,39,39,39,41,43,44,44,44,45,45,45,46,48 , 49,49,49,49,50]
A = solution ()
Print "Result:%d"% (A.removeduplicates (nums))
Print Nums

Implementation is achieved, but with more than 500 ms~~. There's still a lot of room for improvement!

Baidu a bit, found himself guilty again, a little change, you can improve the efficiency of a wide range:

def removeduplicates (self, nums):
#total_length = Len (nums)
If Len (nums) = = 0:return None
Result=1
For index in range (0,len (nums)-1):
#判断后面的元素不等于前面的元素, Count +1, and assign to the previous segment of the array
If nums[index+1]! = Nums[index]:
Nums[result] = nums[index+1]
Result=result+1
return result

76ms OH:

Remove Duplicates from Sorted Array