LeetCode1209. 删除字符串中的所有相邻重复项 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.

题目:给你一个字符串 s,「k 倍重复项删除操作」将会从 s 中选择 k 个相邻且相等的字母,并删除它们,使被删去的字符串的左侧和右侧连在一起。你需要对 s 重复进行无限次这样的删除操作,直到无法继续为止。在执行完所有删除操作后,返回最终得到的字符串。

思路:双指针。同LeetCode1047. 删除字符串中的所有相邻重复项

工程代码下载 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:栈。栈中的每个元素是{已经出现的次数,对应的字符}。参考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;
    }
};
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