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.

class Solution {
    public String removeDuplicates(String s, int k) {
        Stack<Integer> s2 = new Stack();
        Stack<Character> s1 = new Stack();
        for(char c : s.toCharArray()) {
            if(!s1.isEmpty() && s1.peek() == c) s2.push(s2.peek() + 1);
            else s2.push(1);
            s1.push(c);
            if(s2.peek() == k) {
                for(int i = 0; i < k; i++) {
                    s1.pop();
                    s2.pop();
                }
            }
        }
        StringBuilder sb = new StringBuilder();
        while(!s1.isEmpty()) sb.append(s1.pop());
        return sb.reverse().toString();
    }
}

妹想到 俩stack就能做出来了，一个存当前的char，另一个存当前c存在的长度，如果长度和k相等就一起开始pop。