Question: https://leetcode.com/problems/remove-duplicates-from-sorted-array/description/

Basically, duplicates in a sorted array needs to be removed without creating a new one. I have tested my solution, and it falters when there are 3, 5, 7 (and so on) number of duplicates of the same number. Please give me some insights. I don't know if this is a good place to ask this question, or anywhere else to ask this. Any more related advice is appreciated. Thank you.

Code:

def removeDuplicates(nums):
    i = 0 # moving pointer
    end = len(nums)-1 # points to the end of the array
    # iterates over the array
    while i != end:
        # if current number is equal to the next
        if (nums[i] == nums[i 1]):
            # pop it off
            nums.pop(i)
            # decrement the length of array
            end -= 1
        i  = 1 # increments the moving pointer
    return i   1


nums = [0,0,1,1,1,2,2,3,3,4]
print(removeDuplicates(nums))

CodePudding user response：

The way to remove duplicates from a sorted list in O(1) space and O(n) time is to pack the unique elements at the beginning of the list, and then truncate the list:

def removeDuplicates(nums):
    #new length
    newlen = 0
    for val in nums:
        if newlen == 0 or nums[newlen-1] != val:
            # no matter what, we've already retrieved nums[newlen],
            # so this write doesn't interfere with the rest of the loop
            nums[newlen] = val
            newlen  = 1
    
    # delete the unused space at the end of the list, all at once
    if newlen < len(nums):
        del nums[newlen:]
    return newlen


nums = [0,0,1,1,1,2,2,3,3,4]
print(removeDuplicates(nums))
print(nums)