395. Longest Substring with At Least K Repeating Characters

Find the length of the longest substring T of a given string (consists of lowercase letters only) such that every character in T appears no less than k times.

Example 1:

Input:s = "aaabb", k = 3Output:3The longest substring is "aaa", as 'a' is repeated 3 times.

Example 2:

Input:s = "ababbc", k = 2Output:5The longest substring is "ababb", as 'a' is repeated 2 times and 'b' is repeated 3 times.

自己想了两种方法但都在一个很长的字符串上超时了，

首先是递归方法，先贪心算法找到从开始字母计算的最长字符串，然后将后面的字符串进行递归，返回的结果进行对比紧返回最大的那个值。

class Solution {public:    int longestSubstring(string s, int k) {        int size = s.size();        if(size==0) return 0;        vector<int> v(26,0);        int flag=0;        int max=0;        for(int i = 0;i < size;i++){            v[s[i]-'a']++;            flag=1;            for(int j=0;j<26;j++){                if(v[j]<k&&v[j]>0) flag=0;            }            if(flag==1) max=i+1;        }        //cout<<s<<' '<<max<<' '<<s.substr(max,size-max)<<endl;        int ma = Solution::longestSubstring(s.substr(max>0?max:1,size-1),k);        return ma>max ? ma:max;    }};

也是贪心算法，每一遍循环在上一次循环得到的起点开始，得到最长字符串后，将下一位置也即下一循环的起点算出。

class Solution {public:    int longestSubstring(string s, int k) {        int size = s.size();        if(size==0) return 0;        vector<int> v(26,0);        int flag=0;        int max=0;        int now=0;        int now_i=0;        int i = 0;        while(now<=size){            now_i=0;           for(i = now;i < size;i++){              v[s[i]-'a']++;              flag=1;              for(int j=0;j<26;j++){                if(v[j]<k&&v[j]>0) flag=0;              }              if(flag==1){                now_i=i;                max=i+1-now>max ? i+1-now : max;              }           }           if(now_i!=0&&now_i!=now)  now=now_i++;           else now++;           for(int j=0;j<26;j++){                v[j]=0;           }        }        return max;    }};