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 }
原文地址：https://www.cnblogs.com/Dylan-Java-NYC/p/12324677.html