如何从循环中的可变字符串中删除字符？

Question

我试图在循环中删除可变字符串中的一些字符：

fn main() {
    let mut var1 = String::from("BASDFSADLfasLOONBASDFSADLfasLOON");
    let chars: [char; 5] = ['B', 'A', 'L', 'O', 'N'];
    find_balloon(&mut var1, &chars);
}

fn find_balloon<'a>(s: &'a mut String, chars: &'a [char; 5]) {
    for (i, p) in s.char_indices() {
        if chars.iter().any(|&x| x == p) {
            let mut x = s.remove(i);
            println!("{}", x);
        }
    }
}

我得到的错误如下：

error[E0502]: cannot borrow `*s` as mutable because it is also borrowed as immutable
  --> src/main.rs:10:25
   |
8  |     for (i, p) in s.char_indices() {
   |                   ----------------
   |                   |
   |                   immutable borrow occurs here
   |                   immutable borrow later used here
9  |         if chars.iter().any(|&x| x == p) {
10 |             let mut x = s.remove(i);
   |                         ^^^^^^^^^^^ mutable borrow occurs here

Answer 1

您不能从当前迭代的字符串中删除元素。相反，只需使用迭代器的 filter 方法并收集到一个字符串中：

// using a hashset instead of a slice due to hashsets being faster
// the slice.iter().any approach still works though
use std::collections::HashSet;

fn find_balloon(s: &mut String, chars: &HashSet<char>) {
    let filtered: String = s.chars().filter(|x| !chars.contains(x)).collect();
    println!("{}", filtered);
}

fn main() {
    let mut var1 = String::from("BASDFSADLfasLOONBASDFSADLfasLOON");
    let chars: [char; 5] = ['B', 'A', 'L', 'O', 'N'];
    find_balloon(&mut var1, &chars.iter().cloned().collect());
}

Answer 2

您可以使用 String::retain 方法。它完全符合您的要求，无需额外分配。

fn find_balloon(s: &mut String, chars: &[char; 5]) {
    s.retain(|p| if chars.contains(&p) {
        false
    } else {
        print!("{}", p);
        true
    });
}

fn main() {
    let mut var1 = String::from("BASDFSADLfasLOONBASDFSADLfasLOON");
    let chars: [char; 5] = ['B', 'A', 'L', 'O', 'N' ];
    find_balloon(&mut var1, &chars);
}

如何从循环中的可变字符串中删除字符？

How do I remove characters from a mutable string within a loop?

rust

mutable