LeetCode每日一题(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*code”
Output: “lecoe”

Explanation: Performing the removals from left to right:

• The closest character to the 1st star is ‘t’ in “leet*code”. s becomes “leecode”.
• 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.

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用一个 max heap 来保存已遍历过的非*字符的位置， 遍历时如果遇到*字符就从 max heap 中取出一个字符，将两个字符同时消除掉， 如果 max heap 为空则无法消除

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use std::collections::BinaryHeap;

impl Solution {
    pub fn remove_stars(s: String) -> String {
        let mut non_star_char_positions = BinaryHeap::new();
        let mut chars: Vec<char> = s.chars().collect();
        for i in 0..chars.len() {
            if chars[i] != '*' {
                non_star_char_positions.push(i);
                continue;
            }
            if let Some(j) = non_star_char_positions.pop() {
                chars[j] = '-';
                chars[i] = '-';
            }
        }
        chars.into_iter().filter(|&c| c != '-').collect()
    }
}