LeetCode 1209. Remove All Adjacent Duplicates in String II 有坑

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该博客讨论了一种字符串处理问题，即每次选择k个相邻且相等的字符进行删除，直到无法继续删除。给出了一种解决方案，通过维护一个记录字符及其出现次数的列表，逐步删除满足条件的字符，并返回最终结果。示例展示了不同输入和k值下的字符串变化过程，强调了算法的执行逻辑。

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.

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有小坑，注意是每次==k而不是>=k

class Solution:
    def removeDuplicates(self, s: str, k: int) -> str:
        sta = []
        for ch in s:
            if (sta and sta[-1][1] == k):
                sta.pop()
            if (sta and sta[-1][0] == ch):
                sta[-1][1] += 1
            else:
                sta.append([ch,1])
        if (sta and sta[-1][1] == k):
            sta.pop()
        res = ''
        for ch,cnt in sta:
            res += ch*cnt
        return res