Longest Substring with At Least K Repeating Characters (M)
題目
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
題意
在指定字串中找到一個最長的子串，使其包含的每個字母出現的次數都大于等于一個閾值，
思路
求子串問題一般都會想到滑動視窗，但因為本題是要求超過閾值，直接用滑動視窗很難想出一個用來判斷該縮短還是伸長視窗的標準，官方解答提供了一個很巧妙的思路：一個子串能擁有的不同字符的種類是受原字串限制的，所以可以將視窗變動的依據定為“當前子串擁有的不同字符的個數”，這樣最多進行26次遍歷即可得到答案，
代碼實作
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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