LeetCode 1209. Remove All Adjacent Duplicates in String II

原题链接在这里：https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/

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.

题解：

Have two points i and j.

j points to current index in s.

i points to end of result.

Array count is the number of adjacent duplicates.

first do arr[i] = arr[j].

Update count[i]. If arr[i] == arr[i - 1], then count[i] = count[i - 1] + 1.

When count[i] hits k, then set back i by k.

Time Complexity: O(n). n = s.length().

Space: O(n).

AC Java:

class Solution {
    public String removeDuplicates(String s, int k) {
        if(s == null || s.length() < k){
            return s;
        }
        
        int n = s.length();
        int [] count = new int[n];
        int i = 0;
        char [] arr = s.toCharArray();
        for(int j = 0; j < n; j++, i++){
            arr[i] = arr[j];
            count[i] = i > 0 && arr[i] == arr[i - 1] ? count[i - 1] + 1 : 1;
            if(count[i] == k){
                i -= k;
            }
        }
        
        return new String(arr, 0, i);
    }
}

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