LeetCode1209. 刪除字符串中的所有相鄰重複項 II

1209. Remove All Adjacent Duplicates in String II

[Medium] 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.

---

題目：給你一個字符串 s，「k 倍重複項刪除操作」將會從 s 中選擇 k 個相鄰且相等的字母，並刪除它們，使被刪去的字符串的左側和右側連在一起。你需要對 s 重複進行無限次這樣的刪除操作，直到無法繼續爲止。在執行完所有刪除操作後，返回最終得到的字符串。

思路：雙指針。同LeetCode1047. 刪除字符串中的所有相鄰重複項。

工程代碼下載 Github

class Solution {
public:
    string removeDuplicates(string s, int k) {
        int i = 0, n = s.size();
        vector<int> cnt(n);
        for (int j = 0; j < n; ++j, ++i) {
            s[i] = s[j];
            cnt[i] = (i > 0 && s[i-1] == s[j]) ? cnt[i-1] + 1 : 1;
            if (cnt[i] == k)
                i = i - k;
        }
        return s.substr(0, i);
    }
};

思路2：棧。棧中的每個元素是{已經出現的次數，對應的字符}。參考lee215。

class Solution {
public:
    string removeDuplicates(string s, int k) {
        vector<pair<int, char>> sk = {{0, '#'}};
        for(auto c : s){
            if(sk.back().second != c)
                sk.push_back({1, c});
            else if(++sk.back().first == k)
                sk.pop_back();
        }

        string res;
        for(auto p : sk){
            res += string(p.first, p.second);
        }

        return res;
    }
};