[LeetCode] 2390. Removing Stars From a String

You are given a string s, which contains stars *.

In one operation, you can:
Choose a star in s.
Remove the closest non-star character to its left, as well as remove the star itself.
Return the string after all stars have been removed.

Note:
The input will be generated such that the operation is always possible.
It can be shown that the resulting string will always be unique.

Example 1:
Input: s = "leet**cod*e"
Output: "lecoe"
Explanation: Performing the removals from left to right:

• The closest character to the 1st star is 't' in "leet**code". s becomes "leecod*e".
• The closest character to the 2nd star is 'e' in "leecode". s becomes "lecod*e".
• The closest character to the 3rd star is 'd' in "lecod*e". s becomes "lecoe".
  There are no more stars, so we return "lecoe".

Example 2:
Input: s = "erase*****"
Output: ""
Explanation: The entire string is removed, so we return an empty string.

Constraints:
1 <= s.length <= 105
s consists of lowercase English letters and stars *.
The operation above can be performed on s.

从字符串中移除星号。

给你一个包含若干星号 * 的字符串 s 。

在一步操作中，你可以：

选中 s 中的一个星号。
移除星号 左侧 最近的那个 非星号 字符，并移除该星号自身。
返回移除 所有 星号之后的字符串。

注意：

生成的输入保证总是可以执行题面中描述的操作。
可以证明结果字符串是唯一的。

来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/removing-stars-from-a-string
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

思路

思路是 stack。如果遇到字母则入栈；遇到星号的时候，如果栈内有字母则弹出，没有字母则可以直接丢弃星号。

复杂度

时间O(n)
空间O(n) - stack

代码

Java实现

class Solution {
    public String removeStars(String s) {
        Deque<Character> queue = new ArrayDeque<>();
        for (char c : s.toCharArray()) {
            if (c == '*' && !queue.isEmpty()) {
                queue.pollLast();
            } else {
                queue.offerLast(c);
            }
        }

        StringBuilder sb = new StringBuilder();
        while (!queue.isEmpty()) {
            sb.append(queue.pollFirst());
        }
        return sb.toString();
    }
}