[LeetCode] 395. Longest Substring with At Least K Repeating Characters 至少有K个重复字符的最长子字符串

2021年09月15日 阅读数:1
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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.html

Example 1:java

Input:
s = "aaabb", k = 3

Output:
3

The longest substring is "aaa", as 'a' is repeated 3 times.

Example 2:python

Input:
s = "ababbc", k = 2

Output:
5

The longest substring is "ababb", as 'a' is repeated 2 times and 'b' is repeated 3 times.

Java:app

public int longestSubstring(String s, int k) {
    HashMap<Character, Integer> counter = new HashMap<Character, Integer>();
 
    for(int i=0; i<s.length(); i++){
 
        char c = s.charAt(i);
        if(counter.containsKey(c)){
            counter.put(c, counter.get(c)+1);
        }else{
            counter.put(c, 1);
        }
 
    }
 
    HashSet<Character> splitSet = new HashSet<Character>();
    for(char c: counter.keySet()){
        if(counter.get(c)<k){
            splitSet.add(c);
        }
    }
 
    if(splitSet.isEmpty()){
        return s.length();
    }
 
    int max = 0;
    int i=0, j=0;
    while(j<s.length()){
        char c = s.charAt(j);
        if(splitSet.contains(c)){
            if(j!=i){
                max = Math.max(max, longestSubstring(s.substring(i, j), k));
            }
            i=j+1;
        }
        j++;
    }
 
    if(i!=j)
         max = Math.max(max, longestSubstring(s.substring(i, j), k));
 
    return max;
}  

Python:less

class Solution(object):
    def longestSubstring(self, s, k):
        """
        :type s: str
        :type k: int
        :rtype: int
        """
        def longestSubstringHelper(s, k, start, end):
            count = [0] * 26
            for i in xrange(start, end):
                count[ord(s[i]) - ord('a')] += 1
            max_len = 0
            i = start
            while i < end:
                while i < end and count[ord(s[i]) - ord('a')] < k:
                    i += 1
                j = i
                while j < end and count[ord(s[j]) - ord('a')] >= k:
                    j += 1

                if i == start and j == end:
                    return end - start

                max_len = max(max_len, longestSubstringHelper(s, k, i, j))
                i = j
            return max_len

        return longestSubstringHelper(s, k, 0, len(s))

C++:post

class Solution {
public:
    int longestSubstring(string s, int k) {
        int res = 0, i = 0, n = s.size();
        while (i + k <= n) {
            int m[26] = {0}, mask = 0, max_idx = i;
            for (int j = i; j < n; ++j) {
                int t = s[j] - 'a';
                ++m[t];
                if (m[t] < k) mask |= (1 << t);
                else mask &= (~(1 << t));
                if (mask == 0) {
                    res = max(res, j - i + 1);
                    max_idx = j;
                }
            }
            i = max_idx + 1;
        }
        return res;
    }
};

C++:url

class Solution {
public:
    int longestSubstring(string s, int k) {
        int n = s.size(), max_idx = 0, res = 0;
        int m[128] = {0};
        bool ok = true;
        for (char c : s) ++m[c];
        for (int i = 0; i < n; ++i) {
            if (m[s[i]] < k) {
                res = max(res, longestSubstring(s.substr(max_idx, i - max_idx), k));
                ok = false;
                max_idx = i + 1;
            }
        }
        return ok ? n : max(res, longestSubstring(s.substr(max_idx, n - max_idx), k));
    }
};

  

 

 

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