1209. Remove All Adjacent Duplicates in String II (M)

Remove All Adjacent Duplicates in String II (M)

题目

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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题意

给定一个字符串s和整数k，遍历s，如果遇到k个连续字符都相同，则删去这k个字符，并得到新字符串t，对t进行同样的操作，直到最后无字符可删。返回最终得到的字符串。

思路

维护两个栈，分别保存字符和其对应的连续的次数。遍历字符串，如果当前字符与前一个不同，将前一个字符和对应次数压栈；如果当前字符与前一个相同，连续次数累加。然后判断连续次数是否已达到k，如果是则出栈前一个字符和对应次数(相当于将当前字符抛弃)。注意遍历到最后一个字符串时要进行特殊处理。

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代码实现

Java

class Solution {
    public String removeDuplicates(String s, int k) {
        Deque<Character> chars = new ArrayDeque<>();
        Deque<Integer> cnts = new ArrayDeque<>();

        int cnt = 0;
        Character pre = null;
        for (int i = 0; i < s.length(); i++) {
            char cur = s.charAt(i);
            if (pre == null) {
                pre = cur;
                cnt = 1;
            } else if (cur == pre) {
                cnt++;
            }else {
                chars.push(pre);
                cnts.push(cnt);
                pre = cur;
                cnt = 1;
            }

            if (i == s.length() - 1 && cnt < k) {
                chars.push(pre);
                cnts.push(cnt);
            } else if (i < s.length() - 1 && cnt == k) {
                pre = chars.isEmpty() ? null : chars.pop();
                cnt = cnts.isEmpty() ? 0 : cnts.pop();
            }
        }

        StringBuilder sb = new StringBuilder();
        while (!chars.isEmpty()) {
            sb.append((chars.removeLast() + "").repeat(cnts.removeLast()));
        }

        return sb.toString();
    }
}