o(n)遍历一遍即可。
#include<bits/stdc++.h> #define int long long #define double long double #define x first #define y second using namespace std; typedef long long LL; typedef long long ll; typedef pair<int, int> PII; const int N = 3e5 + 10; const int M = 1e3 + 10; int mod = 1e9 + 7; int a[N]; void solve() { int n; cin >> n; for (int i = 1; i <= n; i++) cin >> a[i]; int maxx = 0, cnt = 0; a[n + 1] = 1e18; for (int i = 1; i <= n; i++) { if (abs(a[i] - a[i + 1]) > 1) { maxx = max(maxx, cnt + 1); cnt = 0; } else cnt++; } cout << maxx << "\n"; } signed main() { ios::sync_with_stdio(false); cin.tie(0), cout.tie(0); int _; _ = 1; //cin>>_; while (_--) { solve(); } }