Function anagram_solution4

publication/code/chapter03/anagram_solution4.rs:3–33

(s1: &str, s2: &str)

// anagram_solution4.rs

fn anagram_solution4(s1: &str, s2: &str) -> bool {
    if s1.len() != s2.len() {
        return false;
    }

    // 大小为 26 的集合，用于将字符映射为 ASCII 值
    let mut c1 = [0; 26];
    let mut c2 = [0; 26];
    for c in s1.chars() {
        // 97 为字母 a 的 ASCII 值
        let pos = (c as usize) - 97;
        c1[pos] += 1;
    }
    for c in s2.chars() {
        let pos = (c as usize) - 97;
        c2[pos] += 1;
    }

    // 逐个比较 ascii 值
    let mut pos = 0;
    let mut is_anagram = true;
    while pos < 26 && is_anagram {
        if c1[pos] == c2[pos] {
            pos += 1;
        } else {
            is_anagram = false;
        }
    }

    is_anagram
}

fn main() {
    let s1 = "rust";