提供O(n^2)写法，O(n logn)写法，O(n)写法

O(n^2)写法

#include<iostream>
#include<cmath>
using namespace std;
const int N=1010;
int n;
int q[N];
int main()
{
    cin>>n;

    for(int i=0;i<n;i++) cin>>q[i];

    int res=0;
    for(int i=0;i<n;i++)
        for(int j=0;j<n;j++)
        {
            if(abs(q[i]-q[j])==1&&i<j)
                res++;
        }

    cout<<res<<endl;
    return 0;
}

O(nlogn)写法

#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
const int N = 1010;
int n;
int a[N];
bool st[10010];
int main()
{
    scanf("%d", &n);

    for (int i = 0; i < n; i ++ ) scanf("%d", &a[i]);

    sort(a, a + n);

    int res = 0;
    for (int i = 0; i + 1 < n; i ++ )
        if (a[i + 1] - a[i] == 1)
            res ++;

    printf("%d", res);

    return 0;
}

O(n)写法

#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
const int N = 1010;
int n;
int a[N];
bool st[10010];
int main()
{
    scanf("%d", &n);

    for (int i = 0; i < n; i ++ )
    {
        int x;
        scanf("%d", &x);
        a[i] = x;
        st[x] = true;
    }

    int res = 0;
    for (int i = 0; i < n; i ++ )
    {
        int x = a[i], y = x + 1;
        if(st[y]) res++;
    }

    printf("%d", res);

    return 0;
}