再次请教一个关于数组的方程！！

[pitachiocrepe]

题目是
在一个数组中，如果有几个相同的元素连续排在一起，那么只保留一个，返回值是最后数组中的元素个数。
原题如下：
int removeDups(string a[], int n);
For every sequence of consecutive identical items in a, retain only one item of that sequence. Suppose we call the number of retained items r. Then when this functions returns, elements 0 through r-1 of a must contain the retained items (in the same relative order they were in originally), and the remaining elements may have whatever values you want. Return the number of retained items. Here's an example:
string d[9] = {
"lois", "chris", "peter", "peter", "meg", "meg", "meg", "peter", "peter"
};
int p = removeDups(d, 9); // returns 5
// d[0] through d[4] now contain "lois" "chris" "peter" "meg" "peter"
// We no longer care what strings are in d[5] and beyond.

我的程序是：
int removeDups(string a[], int n)
{
if (n<0)
return -1;
else
{
string item;
for (int i=0; i<n-1; i++)
{
item = a[i];
if (item == a[i+1])
{
for (int j=i+1; j<n; j++)
{
a[j-1]=a[j];
}
i--;
n--;
}
}
return n;
}
}
请大神指教哪里错了？？

[pitachiocrepe]

求助求助！！！！！1

[「已注销」]

// 代码没有错

#include <stdlib.h>
#include <stdio.h>
#include <string>
using namespace std;


int removeDups(string a[], int n)
{
    if (n<0)
        return -1;
    else
    {
        string item;
        for (int i=0; i<n-1; i++)
        {
            item = a[i];
            if (item == a[i+1])
            {
                for (int j=i+1; j<n; j++)
                {
                    a[j-1]=a[j];
                }
                i--;
                n--;
            }
        }
        return n;
    }
}

int main(){
	string d[9] = {"lois", "chris", "peter", "peter", "meg", "meg", "meg", "peter", "peter"};
	int p = removeDups(d, 9);  //  returns 5
	printf("%d",p);
	
	return 0;
}
/*
output:
9
  */