原题链接在这里：https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/
题目：
Given a string s, a k duplicate removal consists of choosing k adjacent and equal letters from s and removing them causing the left and the right side of the deleted substring to concatenate together.
We repeatedly make k duplicate removals on s until we no longer can.
Return the final string after all such duplicate removals have been made.
It is guaranteed that the answer is unique.
Example 1:
Input: s = "abcd", k = 2 Output: "abcd" Explanation: There's nothing to delete.
Example 2:
Input: s = "deeedbbcccbdaa", k = 3 Output: "aa" Explanation: First delete "eee" and "ccc", get "ddbbbdaa" Then delete "bbb", get "dddaa" Finally delete "ddd", get "aa"
Example 3:
Input: s = "pbbcggttciiippooaais", k = 2 Output: "ps"
Constraints:
1 <= s.length <= 10^52 <= k <= 10^4sonly contains lower case English letters.
题解：
Follow the instruction, delete the duplicate, when we have a delete, mark changed as true.
While changed is true, continue.
Time Complexity: O(n^2/k - n). n = s.length(). each level, s becomes n - k, totally there are n/k level.
Space: O(n).
AC Java:
1 class Solution { 2 public String removeDuplicates(String s, int k) { 3 if(s == null || s.length() == 0){ 4 return s; 5 } 6 7 if(k == 1){ 8 return ""; 9 } 10 11 boolean changed = true; 12 while(changed){ 13 changed = false; 14 StringBuilder sb = new StringBuilder(); 15 16 int i = 0; 17 while(i < s.length()){ 18 if(i == s.length() - 1 || s.charAt(i) != s.charAt(i + 1)){ 19 sb.append(s.charAt(i)); 20 i++; 21 }else{ 22 int j = i + 1; 23 int count = 1; 24 while(j < s.length() && s.charAt(j) == s.charAt(i) && count < k){ 25 j++; 26 count++; 27 } 28 29 if(count < k){ 30 sb.append(s.substring(i, j)); 31 }else{ 32 changed = true; 33 } 34 35 i = j; 36 } 37 } 38 39 s = sb.toString(); 40 } 41 42 return s; 43 } 44 }