rust 中替代合并字符串的算法-解网

问：

这是我在 rust 中替代合并字符串的代码。我没有得到预期的输出。它适用于等长字符串。问题是当我们有不相等的字符串时，它会跳过一个字符。

impl Solution {
    pub fn merge_alternately(word1: String, word2: String) -> String {
        let mut merged = String::new();
        let mut chars1 = word1.chars();
        let mut chars2 = word2.chars();

        while let (Some(c1), Some(c2)) = (chars1.next(), chars2.next()) {
            merged.push(c1);
            merged.push(c2);
        }
        // Append any remaining characters from word1
        while let Some(c1) = chars1.next() {
            merged.push(c1);
        }

        // Append any remaining characters from word2
        while let Some(c2) = chars2.next() {
            merged.push(c2);
        }
        merged
    }
}

以下是测试用例失败：

输入 word1 = “ab” word2 = “PQRS”

电流输出 “APBQS”

预期输出 “APBQRS”

字符串算法 merge rust-cargo

pub fn merge_alternately(word1: String, word2: String) -> String {
    let mut merged = String::new();
    let mut chars1 = word1.chars();
    let mut chars2 = word2.chars();

    loop {
        match (chars1.next(), chars2.next()) {
            (Some(c1), Some(c2)) => merged.extend([c1, c2]),
            (Some(c1), None) => merged.push(c1),
            (None, Some(c2)) => merged.push(c2),
            (None, None) => break,
        }
    }

    merged
}

fn main() {
    println!("{}", merge_alternately("ab".into(), "pqrs".into()));
}

apbqrs

此外，如果你被允许使用库，那么迭代器的标准库就是 itertools。它有一个 .interleave（） 函数，可以完全满足您的需求：

use itertools::Itertools;

pub fn merge_alternately(word1: String, word2: String) -> String {
    word1.chars().interleave(word2.chars()).collect()
}

fn main() {
    println!("{}", merge_alternately("ab".into(), "pqrs".into()));
}

apbqrs

上一个：按字典顺序打印具有给定字符的所有可能字符串

下一个：使用 Boyer-Moor 算法加速代码

rust 中替代合并字符串的算法

Algorithm for Alternate merge strings in rust

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