//Given a sorted array, remove the duplicates in place such that each element appear only
// once and return the new length.
// Do not allocate extra space for another array, you must do this in place with constant memory.
// For example, Given input array A = [1,1,2],
// Your function should return length = 2, and A is now [1,2].
//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]
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
class Solution
{
public:
int removeDuplicates(int a[], int n)
{
if (n < 2)
{
return n;
}
int p = 0;
int pn = 1;
int flag = 0;
while(pn < n)
{
if (a[p] == a[pn])
{
if(flag <1)
{
p++;
a[p] = a[pn];
pn++;
flag++;
}
else
{
pn++;
flag ++;
}
}
else
{
p++;
a[p] = a[pn];
pn++;
flag = 0;
}
}
return p+1;
}
};
class Solution
{//更简洁的办法
public:
int removeDuplicates(int A[], int n)
{
if(n<2)return n;
int index = 2;
for (int i = 2; i < n; i++)
{
if (A[i] != A[index-2])
{
A[index] = A[i];
index++;
}
}
return index++;
}
};
int main()
{
int a[12] = {1,1,2,2,2,4,4,7,7,7,8,9};
Solution s;
int t = s.removeDuplicates(a,12);
cout<<t<<endl;
for (int i = 0; i<t; i++)
{
cout<<a[i]<<endl;
}
return 0;
}