1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27

Solved
Easy
Topics
Stack | Queue |String
Companies

You are given a string s consisting of lowercase English letters. A duplicate removal consists of choosing two adjacent and equal letters and removing them.

We repeatedly make duplicate removals on s until we no longer can.

Return the final string after all such duplicate removals have been made. It can be proven that the answer is unique.

Example 1:

Input: s = "abbaca"
Output: "ca"
Explanation: 
For example, in "abbaca" we could remove "bb" since the letters are adjacent and equal, and this is the only possible move.  The result of this move is that the string is "aaca", of which only "aa" is possible, so the final string is "ca".
Example 2:

Input: s = "azxxzy"
Output: "ay"
 
Constraints:

1 <= s.length <= 105
s consists of lowercase English letters.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18

 /*
  * 思路：
          solution1: 队列实现
         遍历字符串，并存入栈，
         当当前元素和队列第一个元素相同时，跳过此次循环i++，同时弹出队列第一个元素；
         否则将当前元素存入队列；
         遍历完后，遍历弹出队列元素组成新字符串即可
       
        solution2: in-place 将字符串当栈使用
       
        solution3: 双指针实现
       
        solution4: 栈实现
         遍历字符串，并存入栈，
         当当前元素和栈顶元素相同时，跳过此次循环i++，同时弹出栈顶元素；
         否则将当前元素存入栈；
         遍历完后，遍历弹出栈内元素组成新字符串，并反转字符串即可
 */

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20

import java.sql.Array;
import java.util.ArrayDeque;

class Solution1 {
    public String removeDuplicates(String s) {
        ArrayDeque deque = new ArrayDeque<>();
        for(int i =0;i<s.length();i++){
            if(deque.isEmpty() || (char) deque.peek() != s.charAt(i)){
                deque.push(s.charAt(i));
            } else{
                deque.pop();
            }
        }
        String newS = "";
        while (!deque.isEmpty()) {
            newS = deque.pop()+ newS;  
        }
        return newS;
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22

class Solution {
    public String removeDuplicates(String s) {
        char[] charS = s.toCharArray();
        int fast = 0;
        int slow = 0;
        
        while (fast < charS.length) {  
            // Use the fast pointer to overwrite the value at the slow pointer
            
            // When encountering the same value before and after, the slow pointer retreats, 
            // and the same value will be overwritten by the fast pointer in the next loop
            if(slow > 0 && charS[slow-1] == charS[fast]){
                slow--;
            } else{
                charS[slow++] = charS[fast]; 
            } 
            //每次循环都需要移动fast
            fast++;
        }
        return new String(charS,0,slow);
    }
}