LeetCode每日一题(1209. Remove All Adjacent Duplicates in String II)

You are given a string s and an integer k, 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 <= 105
• 2 <= k <= 104
• s only contains lower case English letters.

---

这题以前做过， 但是这次提交超时了， 看了看，上次能通过纯属侥幸， 于是换了种方法重新写了一下。
我们创建一个 stack 用来保存入栈的字符和数量, 遍历整个字符串， 对每个字符进行操作， 如果 stack 里处于栈顶的字符跟当前字符相同，则数量加 1， 如果数量达到了 k，则需要弹栈。其他情况直接将当前字符和数量 1 压入栈中即可。

---

impl Solution {
    pub fn remove_duplicates(s: String, k: i32) -> String {
        let mut stack: Vec<(char, i32)> = Vec::new();
        for c in s.chars() {
            if let Some((pc, mut count)) = stack.pop() {
                if pc == c {
                    count += 1;
                    if count != k {
                        stack.push((pc, count));
                    }
                    continue;
                }
                stack.push((pc, count));
                stack.push((c, 1));
                continue;
            }
            stack.push((c, 1));
        }
        stack.into_iter().map(|(c, count)| c.to_string().repeat(count as usize)).fold(String::new(), |mut p, s| {
            p.push_str(&s);
            p
        })
    }
}