思路：我们按照从小到大的顺序将数组逆转好，然后枚举数组首项，分别让其+1，-1，+0，然后求出公差，从前往后遍历即可。

代码：

int ans1(){//不动int cha = (a[n] - a[1] + 1) / (n - 1);int t = a[1];int cnt = 0;for(int i = 2;i <= n;i ++){t += cha;if(abs(t - a[i]) > 1)return 0x3f3f3f3f;if(t != a[i])cnt ++;}return cnt;
}int ans2(){//+1int cha = (a[n] - a[1]) / (n - 1);int t = a[1] + 1;int cnt = 1;for(int i = 2;i <= n;i ++){t += cha;if(abs(t - a[i]) > 1)return 0x3f3f3f3f;if(t != a[i])cnt ++;}return cnt;
}int ans3(){//-1int cha = (a[n] - a[1] + 2) / (n - 1);int t = a[1] - 1;int cnt = 1;for(int i = 2;i <= n;i ++){t += cha;if(abs(t - a[i]) > 1)return 0x3f3f3f3f;if(t != a[i])cnt ++;}return cnt;
}void solve(){cin >> n;vector<int>b(n + 1);for(int i = 1;i <= n;i ++)cin >> a[i], b[i] = a[i];sort(b.begin() + 1,b.end());if(a[1] > a[n])reverse(a + 1,a + 1 + n);if(n <= 2){cout << 0;return;}int ans = min({ans1(),ans2(),ans3()});if(b[1] == b[n])ans = 0;if(ans > n)cout << -1;elsecout << ans;
}