Remove duplicates from Sorted Array II

Remove duplicates from Sorted Array II

Follow up for "Remove duplicates":
What if duplicates is allowed at the most twice?

For example,
Given sorted Array A = [1,1,1,2,2,3] ,

Your function should return length = 5 , and A is now [1,1,2,2,3] .

Tags: Array of pointers
Analysis

Add a variable to record the number of occurrences of the element. This is because it is a sorted array, so a variable can be solved. If there is no sorted array, you need to introduce a hashmap to record the number of occurrences.

Code 1

#include <iostream>using namespacestd;intarr[ -];intRemovetwoduplicate (intA[],intN) {    intindex =0; if(N <=2) {Index=N; }    Else{Index=2;  for(inti =2; I < n; i++){            if(Arr[index-2] !=Arr[i]) {Arr[index]=Arr[i]; Index++; }        }    }    returnindex;}intMain () {intN; CIN>>N;  for(inti =0; I < n; i++) {cin>>Arr[i]; }    intA =removetwoduplicate (arr,n); cout<< a <<Endl;  for(inti =0; i < A; i++) {cout<< Arr[i] <<Endl; }}

Code Listing 2:

#include <iostream>using namespacestd;intarr[ -];intMain () {intN; CIN>>N;  for(inti =0; I < n; i++) {cin>>Arr[i]; }    intindex =0; intCount =1; if(N >2 ){        for(intj =1; j< N; J + +){        if(Arr[index]! =Arr[j]) {Index+=1; Arr[index]=Arr[j]; Count=1; }        Else{Count++; if(Count <=2) {Index++; Arr[index]=Arr[j]; }        }    }     for(inti =0; I <= index; i++) {cout<<Arr[i]; } cout<<Endl; cout<< index+1<<Endl; }    Else {       for(inti =0; I < n; i++) {cout<< arr[i]<<Endl; } cout<< N <<Endl; }    return 0;}

Remove duplicates from Sorted Array II