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. ...
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(...
# @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...
输入:s = "a good example" 输出:"example good a" 解释:如果两个单词间有多余的空格,反转后的字符串需要将单词间的空格减少到仅有一个。 提示: 1 <= s.length <= 104 s 包含英文大小写字母、数字和空格 ' ' s 中至少存在一个 单词 进阶:如果字符串在你使用的编程语言中是一种可变数据类型,请尝试...
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...
[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....
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)...
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 ...