51 lines
1.5 KiB
Rust
51 lines
1.5 KiB
Rust
use crate::Solution;
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impl Solution {
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/// 344.反转字符串
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///
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/// [原题链接](https://leetcode-cn.com/problems/reverse-string/)
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///
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/// 2021-07-20 22:32:53
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///
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/// 编写一个函数，其作用是将输入的字符串反转过来。输入字符串以字符数组 `char[]` 的形式给出。
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///
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/// 不要给另外的数组分配额外的空间，你必须原地修改输入数组、使用 `O(1)` 的额外空间解决这一问题。
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///
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/// 你可以假设数组中的所有字符都是 `ASCII` 码表中的可打印字符。
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///
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/// + 示例 1：
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/// - 输入：`["h","e","l","l","o"]`
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/// - 输出：`["o","l","l","e","h"]`
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/// + 示例 2：
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/// - 输入：`["H","a","n","n","a","h"]`
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/// - 输出：`["h","a","n","n","a","H"]`
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///
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/// Related Topics 递归 双指针 字符串
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///
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/// 👍 425 👎 0
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pub fn reverse_string(s: &mut Vec<char>) {
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let n = s.len();
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let mut idx = 0;
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for i in 0..n / 2 {
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idx = n - i - 1;
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s[i] = (s[i] as u8 ^ s[idx] as u8) as char;
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s[idx] = (s[i] as u8 ^ s[idx] as u8) as char;
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s[i] = (s[i] as u8 ^ s[idx] as u8) as char;
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}
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}
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}
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#[cfg(test)]
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mod test {
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use crate::Solution;
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#[test]
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fn test_q0344() {
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let mut s = vec!['h', 'e', 'l', 'l', 'o'];
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Solution::reverse_string(&mut s);
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assert_eq!(s, vec!['o', 'l', 'l', 'e', 'h']);
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}
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}
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