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.

思路：這題有個tricky的點就是：stack.peek()的count必須跟着當前的點，什麼意思，就是bbyybb,你不能說b的count是4了，你就開始pop，不行，後面的b，count是2，而不是4. 這樣的話，就建立node，node包含char和fre，如果 == k, 直接pop這個node就可以了；

class Solution {
    private class Node {
        public char c;
        public int fre;
        public Node(char c, int fre) {
            this.c = c;
            this.fre = fre;
        }
    }
    
    public String removeDuplicates(String s, int k) {
        Stack<Node> stack = new Stack<Node>();
        for(int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            if(stack.isEmpty() || stack.peek().c != c) {
                stack.push(new Node(c, 1));
            } else {
                stack.peek().fre += 1;
                if(stack.peek().fre == k) {
                    stack.pop();
                }
            }
        }
        
        StringBuilder sb = new StringBuilder();
        while(!stack.isEmpty()) {
            int fre = stack.peek().fre;
            char c = stack.peek().c;
            while(fre > 0) {
                sb.insert(0, c);
                fre--;
            }
            stack.pop();
        }
        return sb.toString();
    }
}