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 <= S.length <= 20000
S
consists only of English lowercase letters.
class Solution { public String removeDuplicates(String S) { char[] ch = S.toCharArray(); Stack<Character> stack = new Stack<>(); for(int i = 0; i < ch.length; i++){ if(!stack.isEmpty()){ if(ch[i] == stack.peek()) stack.pop(); else stack.push(ch[i]); } else stack.push(ch[i]); } StringBuilder res = new StringBuilder(); while(!stack.isEmpty()) { res.append(stack.pop()); } return res.reverse().toString(); } }
Stack啊Stack,FILO,空就push,不空就和当前char做对比,如果相同就pop,不同就push
总结:
This question we used stack to maintain the top of chars that already read, then if we find the next unread char == the top of stack, we just pop it and continue.
If they doesn't match each other or stack is empty, we push current char to stack.
Finally we return the reverse of current string.