[LeetCode] 1209. Remove All Adjacent Duplicates in String II

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 <= 10^5
2 <= k <= 10^4
s only contains lower case English letters.

删除字符串中的所有相邻重复项 II。

给你一个字符串 s，「k 倍重复项删除操作」将会从 s 中选择 k 个相邻且相等的字母，并删除它们，使被删去的字符串的左侧和右侧连在一起。

你需要对 s 重复进行无限次这样的删除操作，直到无法继续为止。

在执行完所有删除操作后，返回最终得到的字符串。

本题答案保证唯一。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/remove-all-adjacent-duplicates-in-string-ii
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题意跟版本一差不多，但是这道题不只是删除重复元素，而是如果某个字符重复了 K 次才删除。我这里提供三种做法，大体思路都是需要操作StringBuilder来删除重复元素。

暴力。遍历input字符串，同时用一个count变量记录当前重复字母出现的次数，如果count == K，则去StringBuilder里面删除这一段(i - k + 1, i + 1)。注意删除操作的区间也是左闭右开的。

时间O(n)

空间O(n)

Java实现

 1 class Solution {
 2     public String removeDuplicates(String s, int k) {
 3         StringBuilder sb = new StringBuilder(s);
 4         int len = -1;
 5         while (len != sb.length()) {
 6             len = sb.length();
 7             for (int i = 0, count = 1; i < sb.length(); i++) {
 8                 if (i == 0 || sb.charAt(i) != sb.charAt(i - 1)) {
 9                     count = 1;
10                 } else {
11                     count++;
12                     if (count == k) {
13                         sb.delete(i - k + 1, i + 1);
14                         break;
15                     }
16                 }
17             }
18         }
19         return sb.toString();
20     }
21 }

计数。这里我们需要用到一个额外数组记录到当前位置，出现了多少次重复字母。当发觉count[i] == K的时候，也是去StringBuilder里面删除这一段(i - k + 1, i + 1)，同时i要往回退K步。

时间O(n)

空间O(n)

Java实现

 1 class Solution {
 2     public String removeDuplicates(String s, int k) {
 3         StringBuilder sb = new StringBuilder(s);
 4         int[] count = new int[sb.length()];
 5         for (int i = 0; i < sb.length(); i++) {
 6             if (i == 0 || sb.charAt(i) != sb.charAt(i - 1)) {
 7                 count[i] = 1;
 8             } else {
 9                 count[i] = count[i - 1] + 1;
10                 if (count[i] == k) {
11                     sb.delete(i - k + 1, i + 1);
12                     i = i - k;
13                 }
14             }
15         }
16         return sb.toString();
17     }
18 }

栈stack。遍历当前字符，如果当前字符跟之前一个字符不同，则往栈内push一个1；如果跟之前一个字符相同，则把栈顶元素拿出来 + 1再push回去。此时如果count == K了，还是需要去StringBuilder里面删除这一段(i - k + 1, i + 1)，同时i要往回退K步。

时间O(n)

空间O(n)

Java实现

 1 class Solution {
 2     public String removeDuplicates(String s, int k) {
 3         StringBuilder sb = new StringBuilder(s);
 4         Deque<Integer> stack = new ArrayDeque<>();
 5         for (int i = 0; i < sb.length(); i++) {
 6             if (i == 0 || sb.charAt(i) != sb.charAt(i - 1)) {
 7                 stack.push(1);
 8             } else {
 9                 int count = stack.pop() + 1;
10                 if (count == k) {
11                     sb.delete(i - k + 1, i + 1);
12                     i = i - k;
13                 } else {
14                     stack.push(count);
15                 }
16             }
17         }
18         return sb.toString();
19     }
20 }