2390.Removing Stars From a String

tags: Medium,String,Stack

2390. Removing Stars From a String

題目描述

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.

範例

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**cod*e". s becomes "lee*cod*e".
- The closest character to the 2nd star is 'e' in "lee*cod*e". 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.

解答

Python

class Solution:
    def removeStars(self, s: str) -> str:
        stack = []

        for ch in s:
            if ch != '*':
                stack.append(ch)
            else:
                stack.pop()

        return "".join(stack)

Ron ChenTue, Apr 11, 2023

class Solution:
    def removeStars(self, s: str) -> str:
        new_s = []
        for i in range(s):
            if i != '*':
                new_s.append(i)
            else:
                new_s.pop()

        ret = ''.join(new_s)
        return ret

gpTue, Apr 11, 2023

Javascript

function removeStars(s) {
  const stack = [];
  for (const c of s) {
    if (c !== '*') {
      stack.push(c);
    } else {
      stack.pop();
    }
  }

  return stack.join('');
}

MarsgoatTue, Apr 11, 2023

C#

public string RemoveStars(string s) {
    char[] output = new char[s.Length];
    int len = 0;
    for (int i = 0; i < s.Length; i++) {
        output[len] = s[i];
        len = s[i] switch {
            '*' => len - 1,
            _ => len + 1
        };
    }

    return string.Concat(output.Take(len));
}

JimWed, Apr 12, 2023

Reference

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