Leetcode第557题 反转字符串中的单词 III

给定一个字符串，你需要反转字符串中每个单词的字符顺序，同时仍保留空格和单词的初始顺序。

示例 1:

输入: "Let's take LeetCode contest" 输出: "s'teL ekat edoCteeL tsetnoc"

注意：在字符串中，每个单词由单个空格分隔，并且字符串中不会有任何额外的空格。

解题思路

这道题目 我们只需要 定义两个指针 互相交换,然后从2边慢慢开始往中间靠近 .

双指针2次交换

use std::ops::Add;
impl Solution {
pub fn reverse_words(s: String) -> String {
    if s == "".to_string(){
        return "".to_string()
    }
    let mut pointer_a  = 0 as usize;
    let mut pointer_b = s.len() -1;
    let mut a =  s.into_bytes();
    while pointer_a < pointer_b {
        let tmp = a[pointer_a];
        a[pointer_a] = a[pointer_b];
        a[pointer_b] = tmp;
        pointer_a +=1;
        pointer_b -=1;
    }

    let mut m:String = a.iter().map(|x| *x as char).collect();
    let mut s: Vec<&str> = m.split(" ").collect();
    pointer_b = s.len() -1;
    pointer_a = 0;
    while pointer_a < pointer_b {
        let tmp = s[pointer_a];
        s[pointer_a] = s[pointer_b];
        s[pointer_b] = tmp;
        pointer_a +=1;
        pointer_b -=1;
    }
    s.join(" ")
    }
}

时间复杂度：O(n)

空间复杂度：O(1)

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