欢迎大家阅读参考，如有错误或疑问请留言纠正，谢谢

Remove Duplicates from Sorted Array

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].

//vs2012测试代码
#include<iostream>

using namespace std;

#define n 5

int main()
{
	int A[n];
	for(int i=0; i<n; i++)
		cin>>A[i];

	//要考虑空数组的情况
	if(n==0)
		return 0;

	int start=0;
	for(int i=0; i<n-1; i++)
	{
		if( A[i]!=A[i+1])
		{
			A[start] = A[i];
			start++;
		}
	}
	A[start] = A[n-1];
	start++;

	//for(int j=start; j<n; j++)
	//	A[j]=0;

	cout<<start<<endl;
	for(int j=0; j<start; j++)
		cout<<A[j];

	return start;
}

//方法一：自测accepted
class Solution {
public:
    int removeDuplicates(int A[], int n) {
	//要考虑空数组的情况
	if(n==0)
		return 0;

	int start=0;
	for(int i=0; i<n-1; i++)
	{
		if( A[i]!=A[i+1])
		{
			A[start] = A[i];
			start++;
		}
	}
	A[start] = A[n-1];
	start++;

	//for(int j=start; j<n; j++)
	//	A[j]=0;

	cout<<start<<endl;
	for(int j=0; j<start; j++)
		cout<<A[j];

	return start;
    }
};

//方法二：参考其他
class Solution {
public:
	int removeDuplicates(int A[], int n) {
		// Start typing your C/C++ solution below
		// DO NOT write int main() function
		if(0 == n) return 0;
		int len = 1;
		for (int i = 1; i < n; ++i)
		{
			if(A[i] != A[len-1])
				A[len++] = A[i];
		}
		return len;
	}
};