Remove Duplicates from Sorted Array II

Remove Duplicates from Sorted Array II

Follow up for “Remove Duplicates”:
What if duplicates are allowed at 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].

标签: Array Two Pointers
分析

加一个变量记录一下元素出现的次数即可。这题因为是已经排序的数组，所以一个变量即可解决。如果是没有排序的数组，则需要引入一个hashmap来记录出现次数。

代码1

#include <iostream>

using namespace std;

int arr[100];

int  removetwoDuplicate(int a[],int n){
    int index = 0;
    if( n <= 2){
        index = n;
    }
    else{
        index = 2;
        for (int  i =2; i < n; i++){
            if (arr[index - 2] != arr[i]){
                arr[index] = arr[i];
                index++;
            }
        }
    }
    return index;
}


int main()
{
    int n;
    cin >> n;
    for (int i = 0; i < n; i++){
        cin >> arr[i];
    }
    int a = removetwoDuplicate(arr,n);
    cout << a <<endl;
    for(int i = 0; i < a; i++){
        cout << arr[i] << endl;
    }
}

代码2：

#include <iostream>

using namespace std;

int arr[100];


int main()
{
    int n;
    cin>>n;
    for(int i = 0; i < n; i++){
        cin >> arr[i];
    }
    int index = 0;
    int count = 1;
    if(n > 2 ){
       for (int j = 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(int  i = 0;i <= index; i++){
        cout << arr[i];

    }
    cout << endl;
    cout << index+1<<endl;
    }
    else {
      for(int i = 0; i < n; i++){
        cout << arr[i]<< endl;
      }
      cout << n << endl;
    }

    return 0;
}