395 Longest Substring with At Least K Repeating Characters 至少有K个重复字符的最长子串
Description:
Given a string s and an integer k, return the length of the longest substring of s such that the frequency of each character in this substring is greater than or equal to k.
Example:
Example 1:
Input: s = "aaabb", k = 3
Output: 3
Explanation: The longest substring is "aaa", as 'a' is repeated 3 times.
Example 2:
Input: s = "ababbc", k = 2
Output: 5
Explanation: The longest substring is "ababb", as 'a' is repeated 2 times and 'b' is repeated 3 times.
Constraints:
1 <= s.length <= 10^4
s consists of only lowercase English letters.
1 <= k <= 10^5
题目描述:
找到给定字符串（由小写字符组成）中的最长子串 T ， 要求 T 中的每一字符出现次数都不少于 k 。输出 T 的长度。
示例 :
示例 1:
输入:
s = "aaabb", k = 3
输出:
3
最长子串为 "aaa" ，其中 'a' 重复了 3 次。
示例 2:
输入:
s = "ababbc", k = 2
输出:
5
最长子串为 "ababb" ，其中 'a' 重复了 2 次， 'b' 重复了 3 次。
思路:
分治
递归的终点为字符串为空或者字符串长度小于 k立即返回 0
递归的检查两边的字符串是否满足大于等于 k这个条件
时间复杂度O(n), 空间复杂度O(n)
代码:
C++:
class Solution
{
public:
int longestSubstring(string s, int k)
{
if (k <= 1) return s.size();
if (s.empty() or s.size() < k) return 0;
vector hash(128, 0);
for (char c : s) ++hash[c];
int i = 0;
while (i < s.size() and hash[s[i]] >= k) ++i;
if (i == s.size()) return s.size();
int l = longestSubstring(s.substr(0, i), k);
while (i < s.size() && hash[s[i]] < k) ++i;
int r = longestSubstring(s.substr(i), k);
return max(l, r);
}
};
Java:
class Solution {
public int longestSubstring(String s, int k) {
if (s == null || s.isEmpty() || s.length() < k) return 0;
if (k <= 1) return s.length();
int count[] = new int[128], i = 0;
for (char c : s.toCharArray()) ++count[c];
while (i < s.length() && count[s.charAt(i)] >= k) ++i;
if (i == s.length()) return s.length();
int l = longestSubstring(s.substring(0, i), k);
while (i < s.length() && count[s.charAt(i)] < k) ++i;
int r = longestSubstring(s.substring(i), k);
return Math.max(l, r);
}
}
Python:
class Solution(object):
def longestSubstring(self, s, k):
if not s:
return 0
for c in set(s):
if s.count(c) < k:
return max(self.longestSubstring(t, k) for t in s.split(c))
return len(s)