1047. Remove All Adjacent Duplicates In String - Easy

Given a string S of lowercase 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 is guaranteed the answer is unique.

 

Example 1:

Input: "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".

 

Note:

1. 1 <= S.length <= 20000
2. S consists only of English lowercase letters.

 

用stack和一个pointer (fast)，fast从index=0开始移动：

如果S[f] = stack.top()，fast右移并弹出栈顶元素；如果S[f] != stack.top()，S[f]入栈并fast右移

time: O(n), space: O(n)

class Solution {
    public String removeDuplicates(String S) {
        if(S.length() <= 1) {
            return S;
        }
        char[] chs = S.toCharArray();
        Deque<Character> stack = new ArrayDeque<>();
        int f = 0;  // fast pointer
        while(f < chs.length) {
            if(!stack.isEmpty() && chs[f] == stack.peekFirst()) {
                f++;
                stack.pollFirst();
            } else {
                stack.offerFirst(chs[f]);
                f++;
            }
        }
        int len = stack.size();
        for(int i = len - 1; i >= 0; i--) {
            chs[i] = stack.pollFirst();
        }
        return new String(chs, 0, len);
    }
}

 

optimized, 用two pointers表示stack

time: O(n), space: O(1)

class Solution {
    public String removeDuplicates(String S) {
        if(S.length() <= 1) {
            return S;
        }
        char[] chs = S.toCharArray();
        int s = -1, f = 0;      // s, f - slow pointer, fast pointer
        while(f < chs.length) {
            if(s > -1 && chs[f] == chs[s]) {
                f++;
                s--;
            } else {
                chs[++s] = chs[f];
                f++;
            }
        }
        return new String(chs, 0, s + 1);
    }
}