题解 | 相助(easy)

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 *
 *                              狗头保佑！
 *                  代码无 bug ！思想全对！算法题一发 ac ！
 */
#include<bits/stdc++.h>
#define ll long long
#define vec vector
using namespace std;
ll rll() 
{
    ll x=0,f=1;
    char ch=getchar();
    while(!isdigit(ch)) 
	{
		if(ch=='-')
			f=-1;
		ch=getchar();
	}
    while(isdigit(ch)) 
	{
		x=x*10+(ch-'0');
		ch=getchar();
	}
    return x*f;
}
int main() 
{
    ll n;
    n=rll();
    vec<ll>a(n);for(int i=0;i<n;++i) a[i]=rll();
    if(n==1){cout<<-1;return 0;} //长度为一不可能实现
    else if(a[0]==a[n-1]) {cout<<1;return 0;} //首尾相同可以直接全删了 1次即可
    else
    {
        for(int i=1;i<n-2;++i) if(a[0]==a[i]&&a[n-1]==a[i+1]) {cout<<2;return 0;}//如果首位不相等 那么我们要找到 一段a[i]和头(a[0]）相等的把这一段删掉 然后找一段a[i+1]和尾a[n-1]相等把后面的删了 这样就可以2次解决
        cout<<-1; //如果没找到就输出-1 结束
        return 0;
    }
    return 0;
}