leetcode刷题557 反转字符串中的单词III Reverse Words in a String III(简单) Python 题目大意: 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...
Python Code: # Define a function 'reverse_string_words' that takes a string 'text' as input.# The function splits the input text into lines, reverses the words within each line, and returns the result.defreverse_string_words(text):forlineintext.split('\n'):return(' '.join(line.split...
代码:oj在线测试通过 Runtime: 172 ms 1classSolution:2#@param s, a string3#@return a string4defreverseWords(self, s):5words = s.split('')67iflen(words) < 2:8returns910tmp =""11forwordinwords:12word = word.replace('','')13ifword !="":14tmp = word +""+tmp1516tmp =tmp.str...
1classSolution:2#@param s, a string3#@return a string4defreverseWords(self, s):5Length=len(s)6if(Length==0):7return""8elif(Length==1):9if(s[0]==""):10return""11else:12returns13else:14num=s[::-1]15list=""16flag_start=017tmp=""18foriinxrange(Length):19if(num[i]==""...
How about multiple spaces between two words? Reduce them to a single space in the reversed string. 1.测试版本代码: AI检测代码解析 # -*- coding: cp936 -*- s=" tian zhai xing " print"Before s:",s defreverseWords(s): L=s.split()#单词拆分成列表 ...
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. fromcollectionsimportdequeclassSolution(object):defreverseWords(self,s):""":type s: str:rtype: str"""ifs==''...
1classSolution:2#@param s, a string3#@return a string4defreverseWords(self, s):5ss = s.split("")6ss = filter(lambdax:x!="",ss)7l =len(ss)8foriinrange(l/2):9ss[i],ss[l-1-i] = ss[l-1-i],ss[i]10s = ("").join(ss)11returns ...
You need to reduce multiple spaces between two words to a single space in the reversed string. Follow up: For C programmers, try to solve it in-place in O(1) space. 题目大意 翻转字符串里面的单词。同时去掉多余的空格。 解题方法 字符串操作直接放弃 C++ ,一般都选择 Python。建议大家刷题的时...
考虑几个特殊的情况 1.若字符窜s=" " 2.字符窜s=“a b d e” 3.字符窜s=“ a”然后在s后面+上一个‘ ’,每次遇到s[i]为空格,s[i-1]不为空格的时候为一个单词 class Solution { public: void reverseWords(string &s)...
public String reverseWords(String s) { int sLen = s.length(), k = 0, j = 0;//j记录空格字符前的索引位置 char strs[] = s.toCharArray(), temp;//转为字符数组 for (int i = 0; i < sLen; i++) { if (strs[i] == ' ') j = i - 1;//遇到空格字符j值减1,为截取的字母段...