leetcode 395 Longest Substring with At Least K Repeating Characters

lc395  Longest Substring with At Least K Repeating Characters

思路是按子串中不同字母的个数来计算，1~26

每次计算n个不同字母所能构成的最长子串

记录其中最长子串

单次循环中，用双指针end向后遍历s，begin用来计算子串长度

当见过的字母个数 == n且这些字母出现次数都大于等于k

说明begin~end这个子串符合条件，更新maxLen

若 >，则需要向后移动begin，使其变为==

若<，不需要操作

class Solution {
    public int longestSubstring(String s, int k) {
        if(s.length() == 0 ||s.length() < k)
            return 0;
        int max = 0;
        for(int i=1; i<=26; i++){
            max = Math.max(max, helper(s, k, i));
        }
        
        return max;
    }
    
    private int helper(String s, int k, int abcNum){
        int[] count = new int[128];
        int begin = 0, end = 0;
        int seenChar = 0, noLessThanK = 0;
        int res = 0;
        
        while(end < s.length()){
            if(count[s.charAt(end)]++ == 0)
                seenChar++;
            if(count[s.charAt(end++)] == k)
                noLessThanK++;
            
            /*while(seenChar > abcNum){
                if(count[s.charAt(begin)]-- == k)
                    noLessThanK--;
                if(count[s.charAt(begin++)] == 0)
                    seenChar--;
            }*/
            while(seenChar > abcNum){
                if(count[s.charAt(begin)]-- == 1)
                    seenChar--;
                if(count[s.charAt(begin++)] == k-1)
                    noLessThanK--;
            }
            
            if(seenChar == noLessThanK && seenChar == abcNum)
                res = Math.max(res, end-begin);
        }
        
        return res;
    }
}

法二：

分治法

统计字符出现次数，用哪些次数少于k的字符将字符串隔开，因为任何包含这些字符的子串肯定不满足条件，这些子串就是子问题

class Solution {
    public int longestSubstring(String s, int k) {
        if(s.length() == 0 || k == 0)
            return 0;
        
        int[] count = new int[128];
        int res = 0;
        
        for(char c : s.toCharArray())
            count[c]++;
        
        List<Integer> split = new ArrayList<Integer>();
        
        for(int i=0; i<s.length(); i++){
            if(count[s.charAt(i)] < k)
                split.add(i);
        }
        
        if(split.size() == 0)
            return s.length();
        split.add(0, -1);
        split.add(s.length());
        
        for(int i=1; i<split.size(); i++){
            int start = split.get(i-1) + 1;
            int end = split.get(i);
            int max = longestSubstring(s.substring(start, end), k);
            res = Math.max(res, max);
        }
        return res;
    }
}