Reverse Word in a String_in a given string you should reverse every word,…-优快云博客

本文链接：https://blog.youkuaiyun.com/sophia_own/article/details/38693201

本文详细阐述了如何实现输入字符串中单词的反转，并确保输出字符串中不包含多余的空格和不必要的前导或尾随空格。通过实例演示了算法的具体步骤和应用场景，包括处理连续空格和特殊输入情况。

Given an input string, reverse the string word by word.

For example,
Given s = "the sky is blue", return "blue is sky the".

Clarification:

What constitutes a word? A sequence of non-space characters constitutes a word.
Could the input string contain leading or trailing spaces? Yes. However, your reversed string should not contain leading or trailing spaces.
How about multiple spaces between two words? Reduce them to a single space in the reversed string.

Testcase:

' a b c d ' -> 'd c b a'
' ' -> ''
' ab bb cc da ' -> 'da cc bb ab'

class Solution:
    # @param s, a string
    # @return a string
    def reverseWords(self, s):
        
        label = True
        result = ''
        index = 0
        std = 0
        stp = len(s)
        
        if len(s) == 0:
            return result
            
        while label:            
            if s[len(s) - 1 - index] == ' ':                
                std = len(s) - index               
                if stp - std == 0:
                    index += 1
                else:
                    label2 = True
                    for dummy_index in range(std, stp):
                        
                        if s[dummy_index] != ' ':
                            if len(result) != 0 and label2:
                                result += ' '
                                label2 = False
                            result += s[dummy_index]
                            label2 = False
                    
                    index += 1
                    stp = len(s) - index
            else:
                index += 1
                std = len(s) - index
                
            if index == len(s):
                label2 = True
                for dummy_index in range(std, stp):                        
                    if s[dummy_index] != ' ':
                        if len(result) != 0 and label2:
                            result += ' '
                            label2 = False
                        result += s[dummy_index]
                        label2 = False
                label = False
            
        return result

Incorrect: (This one does need to consider a lot of things. Ignore!)

reverse string to according to its original settings. s->s1
first element:
is non-space -> add to result
is space -> ignore
middle element:
is non-space -> add to result
is space -> if previous element is space, ignore
-> if previous element is not space, add to result
last element:
is non-space -> add to result
is space -> if previous element is space, remove previous element