至少有K个重复字符的最长子串 · znzryb
## 思路讲解
[https://leetcode.cn/problems/longest-substring-with-at-least-k-repeating-characters/description/](https://leetcode.cn/problems/longest-substring-with-at-least-k-repeating-characters/description/)
其实就是双指针。
```cpp
void cal(ll uni){
vll cnt(28);
ll l=0,r=0;
ll cur=0,bigk=0;
while (r<SZ(s)) {
if(cur<=uni){
int idx=s[r]-'a';
if(cnt[idx]==0) ++cur;
cnt[idx]++;
if(cnt[idx]==K) ++bigk;
++r;
}else{
int idx=s[l]-'a';
if(cnt[idx]==K) --bigk;
--cnt[idx];
if(cnt[idx]==0) --cur;
++l;
}
if(cur==uni && bigk==cur){
ans=max(ans,r-l);
}
}
}
```
## AC代码
AC
[https://leetcode.cn/problems/longest-substring-with-at-least-k-repeating-characters/submissions/658772401/](https://leetcode.cn/problems/longest-substring-with-at-least-k-repeating-characters/submissions/658772401/)
::: 源代码
```cpp
// by Gospel_rock
#include <bits/stdc++.h>
#define all(vec) vec.begin(),vec.end()
#define PB push_back
#define EB emplace_back
#define EM emplace
#define FI first
#define SE second
#define pct __builtin_popcountll
#define ctz __builtin_ctzll
#define SZ(a) ((long long) a.size())
#define FOR(i, a, b) for (long long i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (long long i = (a); i >= (b); --i)
#define debug(var) cerr << #var <<":"<<var<<"\n";
#define cend cerr<<"\n-----------\n"
#define lson(var) (var<<1)
#define rson(var) (var<<1|1)
#define fsp(x) fixed<<setprecision(x)
#define minn(vec) *min_element(vec.begin(),vec.end())
#define maxx(vec) *max_element(vec.begin(),vec.end())
using namespace std;
#ifdef LOCAL
template<typename T> void _print(T x) { cerr << x; }
// 可以打印多层数组，但是格式不是很好
template<typename T> void _print(vector<T> v) { cerr << "[ "; for (T i : v) _print(i), cerr << " "; cerr << "]"; }
template<typename T> void _print(set<T> s) { cerr << "{ "; for (T i : s) _print(i), cerr << " "; cerr << "}"; }
template<typename T, typename U> void _print(pair<T, U> p) { cerr << "{"; _print(p.first); cerr << ", "; _print(p.second); cerr << "}"; }
template<typename T, typename U> void _print(map<T, U> m) { cerr << "{ "; for (auto &p : m) _print(p), cerr << " "; cerr << "}"; }
// 打印两层数组，格式好一些
template<typename T> void _print(vector<vector<T>> v){cerr<<"{\n";for (auto i : v) _print(i), cerr << "\n";cerr<<" }";}
#endif
using ll = long long;using ull = unsigned long long;
using DB=double;using LD=long double;
// using i128 = __int128;
using pdd = pair<DB,DB>;using plb = pair<ll,bool>;
using pll = pair<ll,ll>;using vll=vector<ll>;using vb=vector<bool>;
using tll = tuple<ll,ll,ll>;using vii=vector<int>;
using vpll = vector<pll>;using vtll = vector<tll>;
using vdb = vector<DB>;using vc=vector<char>;
using arr3 = array<ll,3> ;using arr2 = array<ll,2>;
using CD=complex<double>;
static constexpr ll MAXN=(ll)1e6+10,INF=(ll)1e17+3; // 1e17+3是素数 1e18+9是素数
static constexpr ll mod=(ll)1e9+7; // 998244353;
static constexpr double eps=1e-8;const double pi=acos(-1.0);
ll N,M,Q,X,K,T,lT;
inline void __();
signed main()
{
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
// cin>>lT;
// while(lT--)
__();
return 0;
}
/*
*/
string s;
ll ans=0;
void cal(ll uni){
vll cnt(28);
ll l=0,r=0;
ll cur=0,bigk=0;
while (r<SZ(s)) {
if(cur<=uni){
int idx=s[r]-'a';
if(cnt[idx]==0) ++cur;
cnt[idx]++;
if(cnt[idx]==K) ++bigk;
++r;
}else{
int idx=s[l]-'a';
if(cnt[idx]==K) --bigk;
--cnt[idx];
if(cnt[idx]==0) --cur;
++l;
}
if(cur==uni && bigk==cur){
ans=max(ans,r-l);
}
}
}
inline void __(){
cin>>s;
cin>>K;
FOR(i,1,26){
cal(i);
}
cout<<ans<<"\n";
}
/*
*/
```
:::
## 心路历程（WA，TLE，MLE……）