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.
Solution: Stack
Time complexity: O(n)
Space complexity: O(n)
C++
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// Author: Huahua class Solution { public: string removeDuplicates(string s, int k) { vector<pair<char, int>> st{{'*', 0}}; for (char c : s) if (c != st.back().first) st.emplace_back(c, 1); else if (++st.back().second == k) st.pop_back(); string ans; for (const auto& p : st) ans.append(p.second, p.first); return ans; } }; |
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