publicString reverseWords(String s) { if(s ==null|| s.length() ==0) returns; intindex =0;//the pointer to traverse intlen = s.length(); Stack<String> stack =newStack<String>();//堆栈,先进后出,顺序存入单词,逆序输出 StringBuilder sBuilder =newStringBuilder();//记录每一个单词 char[...
1publicclassSolution {2publicString reverseWords(String s) {3if(s.equals(""))returns;4String arr[]=s.split(" ");5String new_word="";6for(inti=arr.length-1;i>=0;i--)7{8if(arr[i].equals(""))continue;9new_word+=arr[i]+" ";10}11new_word=new_word.toString().trim();12re...
Output: "s'teL ekat edoCteeL tsetnoc" Note: In the string, each word is separated by single space and there will not be any extra space in the string. 1. 2. 3. 4. 5. 6. 代码 注意空格。。。最后一个单词后面不追加空格 class Solution { public String reverseWords(String s) { Strin...
AC Java: AI检测代码解析 1classSolution {2publicString reverseWords(String s) {3if(s ==null|| s.length() == 0){4returns;5}67intlen =s.length();8char[] charArr =s.toCharArray();910intlo = 0;11for(inti = 0; i<=len; i++){12if(i==len || charArr[i]==' '){13reverse(...
RegisterAttributeJavaTypeParametersAttribute Remarks Returns a comparator that imposes the reverse ordering of the specified comparator. If the specified comparator isnull, this method is equivalent to#reverseOrder()(in other words, it returns a comparator that imposes the reverse of the natural ordering...
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557. Reverse Words in a String III Easy Given a string, you need to reverse the order of characters in each word within a sentence while still preserving whitespace and initial word order. Example 1: Input: "Let's take LeetCode contest" ...
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string.reverse.deleteCharAt(reverse.length()-1);returnreverse.toString();}// Main method to execute the program.publicstaticvoidmain(String[]args){Stringstr1="Reverse words in a given string";// Given input string.// Display the given string.System.out.println("The given string is: "+str...
class Solution { public: void reverseWords(string &s) { int len = s.length(); reverse(s.begin(), s.end()); int wpos = 0; for (int p=0, q=0; p < len; p = q) { while ((p<len) && (s[p] == ' ')) p++; q = p; while ((q<len) && (s[q] != ' ')) q+...