15 voidreverseWords(string &s) { string rs; for(inti = s.length()-1; i >= 0; ) { while(i >= 0 && s[i] ==' ') i--; if(i < 0)break; if(!rs.empty()) rs.push_back(' '); string t; while(i >= 0 && s[i] !=' ') t.push_back(s[i--]); reverse(t.begin(...
Time complexity: O(n) Space complexity: O(n) classSolution{public:voidreverseWords(string &s){reverse(s.begin(), s.end());stringstreamss(s); string str, ret;while(ss >> str) {reverse(str.begin(), str.end());for(charc : str) { ret.push_back(c); } ret.push_back(' '); }...
# @return a string def reverseWords(self, s): if len(s) == 0: return '' words = s.split(' ') i = 0 while i < len(words): if words[i] == '': del words[i] else: i = i + 1 if len(words) == 0: return '' words.reverse() result = '' for item in words: resul...
public class Solution { public String reverseWords(String s) { String[] words = s.trim().split(" +"); int len = words.length; StringBuilder result = new StringBuilder(); for(int i = len -1; i>=0;i--){ result.append(words[i]); if(i!=0) result.append(" "); } return resu...
给你一个字符串 s ,请你反转字符串中 单词 的顺序。 单词 是由非空格字符组成的字符串。s 中使用至少一个空格将字符串中的 单词 分隔开。 返回单词 顺序颠倒且 单词 之间用单个空格连接的结果字符串。 注意:输入字符串 s中可能会存在前导空格、尾随空格或者单词间的多个空格。返回的结果字符串中,单词间应当...
Given an input string, reverse the string word by word. For example, Given s = "the sky is blue", return "blue is sky the". */ #include <stdio.h> #include <string.h> #include <stdlib.h> struct node { char *dest; struct node *Next; ...
Reverse Words in a String https://leetcode.com/problems/reverse-words-in-a-string/ 给定一个String,求出这个String 的字符串反转,反转后的以空格为区分的单词需要保持原顺序 例如String = "the sky is blue" ,反转后的结果为"blue is sky the",附加条件:两个单词之间或者句子的收尾可能有多个空格,需要...
void reverseWords(string &s) { s = removeDuplicateSpace(s); int begin = 0; int end = 0; while(end < s.size()){ if(s[end] == ' '){ swapString(s, begin, end - 1); begin = end+1; end = begin; } else{ end++; } } swapString(s, begin, end - 1)...
[LeetCode] 186. Reverse Words in a String II Given an input string, reverse the string word by word. A word is defined as a sequence of non-space characters. The input string does not contain leading or trailing spaces and the words are always separated by a single space....
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 ...