<jupyter_start><jupyter_text>151. Reverse Words in a String  ContentGiven an input string s, reverse the order of the words.A word is defined as a sequence of non-space characters. The words in s will be separated by at least one space.Return a string of the words in reverse order concatenated by a single space.Note that s may contain leading or trailing spaces or multiple spaces between two words. The returned string should only have a single space separating the words. Do not include any extra spaces.&nbsp;Example 1:Input: s = &quot;the sky is blue&quot;Output: &quot;blue is sky the&quot;Example 2:Input: s = &quot;  hello world  &quot;Output: &quot;world hello&quot;Explanation: Your reversed string should not contain leading or trailing spaces.Example 3:Input: s = &quot;a good   example&quot;Output: &quot;example good a&quot;Explanation: You need to reduce multiple spaces between two words to a single space in the reversed string.Example 4:Input: s = &quot;  Bob    Loves  Alice   &quot;Output: &quot;Alice Loves Bob&quot;Example 5:Input: s = &quot;Alice does not even like bob&quot;Output: &quot;bob like even not does Alice&quot;&nbsp;Constraints:	1 &lt;= s.length &lt;= 104	s contains English letters (upper-case and lower-case), digits, and spaces &39; &39;.	There is at least one word in s.&nbsp;Follow-up:&nbsp;If the string data type is mutable in your language, can&nbsp;you solve it&nbsp;in-place&nbsp;with&nbsp;O(1)&nbsp;extra space?  Difficulty: Medium, AC rate: 25.3% Question Tags:- Two Pointers- String Links: 🎁 [Question Detail](https://leetcode.com/problems/reverse-words-in-a-string/description/) | 🎉 [Question Solution](https://leetcode.com/problems/reverse-words-in-a-string/solution/) | 💬 [Question Discussion](https://leetcode.com/problems/reverse-words-in-a-string/discuss/?orderBy=most_votes) Hints:  Sample Test Case"the sky is blue" ---What's your idea?注重 `O(1)` 的空间复杂度大体思路上是：1. 先整个字符串翻转2. 然后逐个单词翻转细节实现上，在第2步，为了避免多次遍历，需要记录翻转单词后应该放置的位置（`new_word_start`)对于第1步翻转后的字符串的划分，用正则表示是这样的： `([ ]*)([A-Za-z0-9]+)( )`，即「零个或多个前导空格」+「单词」+「一个后置空格」---<jupyter_code>class Solution:
    def reverseWords(self, s: str) -> str:
        l = [' '] + list(s)
        length = len(l)
        
        # 第1步，整个字符串翻转
        left, right = 0, length-1
        while left < right:
            l[left], l[right] = l[right], l[left]
            left += 1
            right -= 1
        
        # 第2步，逐个单词翻转
        new_word_start = 0
        inner_word_start = 0
        while inner_word_start < length:
            # 前导空格的个数
            leading_space_cnt = 0
            while l[inner_word_start+leading_space_cnt] == ' ':
                leading_space_cnt += 1
                if (inner_word_start+leading_space_cnt >= length):
                    return ''.join(l[:new_word_start-1])
            
            # 单词的长度
            word_size = 0
            while l[inner_word_start+leading_space_cnt+word_size] != ' ':
                word_size += 1
            
            # 单词翻转
            swap_cnt = 0
            while swap_cnt < word_size:
                left = new_word_start+swap_cnt
                right = inner_word_start+leading_space_cnt+word_size-1-swap_cnt
                if (left >= right):
                    break
                l[left], l[right] = l[right], l[left]
                swap_cnt += 1
            
            # 新的单词放置位置递增「单词长度+一个空格」的量
            new_word_start += (word_size+1)
            # 下一翻转单词的位置递增「前导空格数+单词长度+一个空格」的量
            inner_word_start += (leading_space_cnt+word_size+1)
            
        return ''.join(l[:new_word_start-1])
s = Solution()

print('>' + s.reverseWords('the sky is blue') + '<')
print('>' + s.reverseWords('  hello world  ') + '<')
print('>' + s.reverseWords('a good   example') + '<')
print('>' + s.reverseWords('  Bob    Loves  Alice   ') + '<')
print('>' + s.reverseWords('Alice does not even like bob') + '<')
import sys, os; sys.path.append(os.path.abspath('..'))
from submitter import submit
submit(151)<jupyter_output><empty_output>
{'fraction_non_alphanumeric': 0.12080536912751678, 'fraction_numerical': 0.012179965200099428, 'mean_word_length': 2.8581016299137105, 'pattern_counts': {'":': 0, '<': 13, '<?xml version=': 0, '>': 13, 'https://': 3, 'lorem ipsum': 0, 'www.': 0, 'xml': 0}, 'pii_count': 0, 'substrings_counts': 0, 'word_list_counts': {'cursed_substrings.json': 0, 'profanity_word_list.json': 0, 'sexual_word_list.json': 0, 'zh_pornsignals.json': 0}}
{'chain_length': '1', 'dir': 'jupyter-structured-clean-dedup', 'id': '357567', 'license': 'MIT', 'n_tokens_mistral': 1447, 'n_tokens_neox': 1438, 'n_words': 351, 'path': '151-200/151.reverse-words-in-a-string.ipynb', 'repo_name': 'Sorosliu1029/LeetCode', 'size': '4023'}