LeetCode 424. Longest Repeating Character Replacement
原题链接在这里:https://leetcode.com/problems/longest-repeating-character-replacement/description/
题目:
Given a string that consists of only uppercase English letters, you can replace any letter in the string with another letter at most k times. Find the length of a longest substring containing all repeating letters you can get after performing the above operations.
Note:
Both the string's length and k will not exceed 104.
Example 1:
Input: s = "ABAB", k = 2 Output: 4 Explanation: Replace the two 'A's with two 'B's or vice versa.
Example 2:
Input: s = "AABABBA", k = 1 Output: 4 Explanation: Replace the one 'A' in the middle with 'B' and form "AABBBBA". The substring "BBBB" has the longest repeating letters, which is 4.
题解:
类似快慢指针维护substring的方法在Minimum Window Substring里有总结.
随着runner向前推进找出最高frequency word的count. substring整个长度减掉最大的count 就是需要调换的字符数, 若大于k就移动walker.
移动walker时不需要更改maxCount, runner-walker-maxCount自会变小.
Why we don't need to update maxCount, here we only need to care the longest substring, it is maxCount+k.
Thus we only need to care when there is new largert maxCount.
Here while only execuates once, move walker once, it already break the while condition.
Time Complexity: O(s.length()).
Space: O(1).
AC Java:
1 class Solution { 2 public int characterReplacement(String s, int k) { 3 int [] map = new int[256]; 4 int walker = 0; 5 int runner = 0; 6 int maxCount = 0; 7 int res = 0; 8 while(runner < s.length()){ 9 maxCount = Math.max(maxCount, ++map[s.charAt(runner++)]); 10 while(runner-walker-maxCount > k){ 11 map[s.charAt(walker++)]--; 12 } 13 14 res = Math.max(res, runner-walker); 15 } 16 return res; 17 } 18 }
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