CodeForces 888C K-Dominant Character(模拟)
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You are given a string s consisting of lowercase Latin letters. Character c is called k-dominant iff each substring of s with length at least k contains this character c.
You have to find minimum k such that there exists at least one k-dominant character.
Input
The first line contains string s consisting of lowercase Latin letters (1?≤?|s|?≤?100000).
Output
Print one number — the minimum value of k such that there exists at least one k-dominant character.
Example
Input
abacaba
Output
2
Input
zzzzz
Output
1
Input
abcde
Output
3
题意:
任意长度至少为k的子串包含有相同的字母。
题解:
从a到z找一遍就好了,寻找同一字母在序列中出现的最大的间隔,最后取间隔最小的字母。
自己写的虽然过了,感觉代码写的很烂。
#include<iostream>
#include<cstring>
#include<string>
using namespace std;
int main()
{
string s;
while(cin>>s)
{
int last[26];
int a[26]={0},len=s.length();
memset(last,-1,sizeof(last));
for(int i=0;i<len;i++)
{
int t=s[i]-'a';
if(last[t]==-1)
{
a[t]=i+1;
last[t]=i;
}
else
{
a[t]=max(a[t],i-last[t]);
last[t]=i;
}
}
int ans=100000;
for(int i=0;i<26;i++)//处理到字符串结尾的间隔
{
if(last[i]!=-1)
{
a[i]=max(a[i],len-last[i]);
ans=min(ans,a[i]);
}
}
cout<<ans<<endl;
}
return 0;
}
参考别人重新写的
#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
int main()
{
string s;
while(cin>>s)
{
int ans=1e6;
for(int i='a';i<='z';i++)
{
int t=0,k=0;
for(int j=0;j<s.length();j++)
{
if(s[j]==i)
t=0;
else
t++;
k=max(k,t);
}
ans=min(ans,k);
}
cout<<ans+1<<endl;
}
return 0;
}
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