151.翻转字符串里的单词

https://leetcode-cn.com/problems/reverse-words-in-a-string/

Java

/*
 * @Author: Goog Tech
 * @Date: 2020-08-04 10:34:22
 * @LastEditTime: 2020-08-04 11:09:04
 * @Description: https://leetcode-cn.com/problems/reverse-words-in-a-string/
 * @FilePath: \leetcode-googtech\#102. Binary Tree Level Order Traversal\Solution.java
 */

/*
 * @lc app=leetcode.cn id=151 lang=java
 *
 * [151] 翻转字符串里的单词
 */

// @lc code=start
class Solution {

    /**
     * 双指针法
     * 
     * 1.倒序遍历字符串s,记录单词左右索引边界left,right.
     * 2.每确定一个单词的边界,则将其添加到单词列表result中.
     * 3.最后将单词列表拼接为字符串并去除头尾空格,将其返回即可.
     */
    public String reverseWords(String s) {
        // 删除字符串的首尾空格
        s = s.trim();
        // 初始化双指针,使得left与right指向字符串尾字符
        int right = s.length() - 1, left = right;
        StringBuilder result = new StringBuilder();
        // 通过移动左指针来检查字符串中是否含有单词
        while(left >= 0) {
            // left指针向左移动: 寻找第一个空格
            while( left >= 0 && s.charAt(left) != ' ') left--;
            // 将单词添加到result中,并在其后添加空格
            result.append(s.substring(left + 1, right + 1) + " ");
            // 跳过单词之间的空格
            while(left >= 0 && s.charAt(left) == ' ') left--;
            // 使得right指向下一个单词的尾字符
            right = left;
        }
        // 将结果转化为字符串并删除头尾空格
        return result.toString().trim();
    }
}

// @lc code=end

Python

'''
Author: Goog Tech
Date: 2020-08-04 10:55:29
LastEditTime: 2020-08-04 11:10:18
Description: https://leetcode-cn.com/problems/reverse-words-in-a-string/
FilePath: \leetcode-googtech\#102. Binary Tree Level Order Traversal\Solution.py
'''

#
# @lc app=leetcode.cn id=151 lang=python
#
# [151] 翻转字符串里的单词
#

# @lc code=start
class Solution(object):

    # 双指针法
    def reverseWords(self, s):
        """
        :type s: str
        :rtype: str
        """
        result = []
        # 去除字符串的首位空格
        s = s.strip()
        # 初始化双节点,使得left与right指向字符串的尾字符
        left = right = len(s) - 1
        # 通过移动左指针来检查字符串中是否含有单词
        while left >= 0:
            # left指针向左移动: 寻找第一个空格
            while left >= 0 and s[left] != ' ': left -= 1
            # 将单词添加到结果列表
            result.append(s[left + 1 : right + 1])
            # 跳过单词之间的空格
            while left >= 0 and s[left] == ' ': left -= 1
            # 使得right指向下个单词的尾字符
            right = left
        # 将结果列表用空格拼接为字符串并返回
        return ' '.join(result)

# @lc code=end