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();
}
}

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