Leetcode[26]-remove Duplicates from Sorted Array

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

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

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 1 and 2 respectively. It doesn ' t matter what are you leave beyond the new length.

idea : You need to add a new variable pos. The initial value is 0. Used to record the position of non-repeating elements. Begins the traversal of the second element of the array, assuming that it is equal to the preceding element, and jumps directly to the next one. Assuming no difference, the value of the array is assigned to the ++pos bit, and then continues to traverse the next;

Code (c + +):

classSolution { Public:intRemoveDuplicates ( vector<int>& Nums) {intn = nums.size ();if(n==0)return 0;intpos =0, i =1; while(I < N) {if(Nums[i] = = nums[i-1]) {i++; }Else{Nums[++pos] = nums[i++]; }} n = pos+1; Nums.resize (n);returnN }};

Make a simple change:

classSolution { Public:intRemoveDuplicates ( vector<int>& Nums) {intn = nums.size ();if(n==0)return 0;intpos =1, i =1;//Start POS from 1         while(I < N) {if(Nums[i] = = nums[i-1]) {i++; }Else{nums[pos++] = nums[i++];//stay consistent here}} n = pos; Nums.resize (n);returnN }};

Python code

 class solution(object):     def removeduplicates(self, nums):        "" : Type Nums:list[int]: Rtype:int " " "n = Len (nums)ifn = =0:return 0pos =1; i =1; count=0         whileI < n:ifNums[i] = = nums[i-1]: i=i+1count=count+1            Else: Nums[pos] = nums[i] pos = pos+1i = i +1         forIinchRange (count): Nums.pop (-1)returnLen (nums)

Leetcode[26]-remove Duplicates from Sorted Array