Longest Substring with At Least K Repeating Characters (M)
题目
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 less than or equal to k.
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 <= 10^4sconsists of only lowercase English letters.1 <= k <= 10^5
题意
在指定字符串中找到一个最长的子串，使其包含的每个字母出现的次数都大于等于一个阈值。
思路
求子串问题一般都会想到滑动窗口，但因为本题是要求超过阈值，直接用滑动窗口很难想出一个用来判断该缩短还是伸长窗口的标准。官方解答提供了一个很巧妙的思路：一个子串能拥有的不同字符的种类是受原字符串限制的，所以可以将窗口变动的依据定为“当前子串拥有的不同字符的个数”，这样最多进行26次遍历即可得到答案。
代码实现
Java
class Solution {
public int longestSubstring(String s, int k) {
int ans = 0;
int uniqueCount = 0;
boolean[] exist = new boolean[26];
for (char c : s.toCharArray()) {
if (!exist[c - 'a']) {
exist[c - 'a'] = true;
uniqueCount++;
}
}
for (int ceil = 1; ceil <= uniqueCount; ceil++) {
int unique = 0, satisfied = 0, start = 0, end = 0;
int[] count = new int[26];
while (end < s.length()) {
if (unique <= ceil) {
int index = s.charAt(end) - 'a';
if (count[index] == 0) unique++;
count[index]++;
if (count[index] == k) satisfied++;
end++;
} else {
int index = s.charAt(start) - 'a';
if (count[index] == k) satisfied--;
count[index]--;
if (count[index] == 0) unique--;
start++;
}
if (unique == ceil && unique == satisfied) {
ans = Math.max(ans, end - start);
}
}
}
return ans;
}
}
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