在Java中用Word反向字符串
我有以下代码逐字反转字符串，我有一个问题，首先，任何人都可以指出如何使它更好的代码？ 第二，如何删除新字符串开头的空格。
String str = "hello brave new world"; tStr.reverseWordByWord(str) public String reverseWordByWord(String str){ int strLeng = str.length()-1; String reverse = "", temp = ""; for(int i = 0; i = 0; j--){ reverse += temp.charAt(j); if((j == 0) && (i != strLeng)) reverse += " "; } temp = ""; } } return reverse; } 此刻的短语变为：
olleh evarb wen dlrow
注意新字符串开头的空格。
不使用split函数代码看起来像：
public static void reverseSentance(String str) { StringBuilder revStr = new StringBuilder(""); int end = str.length(); // substring takes the end index -1 int counter = str.length()-1; for (int i = str.length()-1; i >= 0; i--) { if (str.charAt(i) == ' ' || i == 0) { if (i != 0) { revStr.append(str.substring(i+1, end)); revStr.append(" "); } else { revStr.append(str.substring(i,end)); } end = counter; } counter--; } System.out.println(revStr); }
如果str =“快速的棕色狐狸跳过懒狗！” 它会像“狗！懒散的跳过狐狸棕快速回归”……
private static String Reverse(String str) { char charArray[] = str.toCharArray(); for (int i = 0; i
这是你如何做到的：
StringBuilder result = new StringBuilder(); StringTokenizer st = new StringTokenizer(input, " "); while (st.hasMoreTokens()) { StringBuilder thisToken = new StringBuilder(st.nextToken()); result.append(thisToken.reverse() + " "); } String resultString = result.toString();
我使用StringUtils的方法。 在unit testing中。
@Test public void testReversesWordsAndThenAllCharacters(){ String sentence = "hello brave new world"; String reversedWords = StringUtils.reverseDelimited(sentence, ' '); String reversedCharacters = StringUtils.reverse(reversedWords); assertEquals("olleh evarb wen dlrow", reversedCharacters); }
如果静态导入StringUtils，可以将其内联到：
reverse(reverseDelimited("hello brave new world", ' '))
我要做的第一件事就是将代码翻转的代码分开，然后单独翻转每个单词。 这个内循环：
for(int j = temp.length()-1; j >= 0; j--) { reverse += temp.charAt(j); if((j == 0) && (i != strLeng)) reverse += " "; }
将是一个函数/方法调用。
另外，为了使代码更高效，而不是使用+运算符连接字符串，我会使用字符串缓冲类。 比如StringBuffer或StringBuilder 。
首先，您应该将它与三个函数分离。 第一个使用空格作为分隔符来破坏字符串列表中的大字符串，第二个反转一个没有空格的字符串，以及最后的连接字符串。
当你这样做时，更容易找到导致空间出现的原因。 您已经可以在当前代码中看到，但我不会告诉您：D。
用这样的东西怎么样？
String string="yourWord"; String reverse = new StringBuffer(string).reverse().toString();
您可以使用StringUtils
return StringUtils.reverseDelimitedString(str, " ");
尝试这个。 它考虑了任何类型的标点符号和空格字符。
public String reverseWordByWord(String inputStr) { BreakIterator wordIterator = BreakIterator.getWordInstance(); wordIterator.setText(inputStr); int start = wordIterator.first(); StringBuilder tempBuilder; StringBuilder outBuilder = new StringBuilder(); for (int end = wordIterator.next(); end != BreakIterator.DONE; start = end, end = wordIterator.next()) { tempBuilder = new StringBuilder(inputStr.substring(start, end)); outBuilder.append(tempBuilder.reverse()); } return outBuilder.toString(); }
public class StringReversers { public static void main(String[] args) { String s = new String(revStr("hello brave new world")); String st = new String(revWords("hello brave new world")); System.out.println(s); System.out.println(st); } public static String revStr(String s){ StringBuilder sb = new StringBuilder(); for (int i=s.length()-1; i>=0;i--){ sb.append(s.charAt(i)); } return sb.toString(); } public static String revWords(String str) { StringBuilder sb = new StringBuilder(); String revd = revStr(str); for (String s : revd.split(" ")){ sb.append(revStr(s)); sb.append(" "); } return sb.toString(); } }
public static void reverseByWord(String s) { StringTokenizer token = new StringTokenizer(s); System.out.println(token.countTokens()); Stack stack = new Stack (); while (token.hasMoreElements()) { stack.push(token.nextElement().toString()); } while (!stack.isEmpty()) { System.out.println(stack.pop()); } }
不使用拆分方法的另一种方案
public static String reverseWordsWithoutSplit(String str) { StringBuffer buffer = new StringBuffer(); int length = str.length(); while(length >0) { int wordstart = length -1; while(wordstart >0 && str.charAt(wordstart) != ' '){ wordstart--; } buffer.append(str.substring(wordstart==0?wordstart:wordstart+1, length)); if(wordstart>0) buffer.append(" "); length = wordstart; } return buffer.toString(); }
另一个方案。 该解决方案就位。
在一个字符串中反转单词（单词由一个或多个空格分隔），空格可以在单词之前，即句子开头的空格，结尾等……就地解决它。
public class ReverseWordsInString { public static void main(String[] args) { // TODO Auto-generated method stub char[] sentence = " Hi my name is person!".toCharArray(); System.out.println(ReverseSentence(sentence)); } private static char[] ReverseSentence(char[] sentence) { //Given: "Hi my name is person!" //produce: "iH ym eman si !nosrep" //the obvious naive solution: utilize stringtokenize to separate each word into its own array. reverse each word and insert space between each array print //better solution: drop stringtokenize and use a counter to count how many characters processed before space was hit. // once space hit, then jump back swap characters between counter-1 and start position. O(1) Space if(sentence == null) return null; if(sentence.length == 1) return sentence; int startPosition=0; int counter = 0; int sentenceLength = sentence.length-1; //Solution handles any amount of spaces before, between words etc... while(counter <= sentenceLength) { if(sentence[counter] == ' ' && startPosition != -1 || sentenceLength == counter) //Have passed over a word so upon encountering a space or end of string reverse word { //swap from startPos to counter - 1 //set start position to -1 and increment counter int begin = startPosition; int end; if(sentenceLength == counter) { end = counter; } else end = counter -1; char tmp; //Reverse characters while(end >= begin){ tmp = sentence[begin]; sentence[begin] = sentence[end]; sentence[end] = tmp; end--; begin++; } startPosition = -1; //flag used to indicate we have no encountered a character of a string } else if(sentence[counter] !=' ' && startPosition == -1) //first time you encounter a letter in a word set the start position { startPosition = counter; } counter++; } return sentence; }
}
这是一个讨论这个问题的线程。 我认为使用正则表达式拆分的一个答案是非常聪明的。
https://codereview.stackexchange.com/questions/43838/reverse-a-string-word-by-word
public String reverseWordByWord(String s) { StringBuilder result = new StringBuilder(); String[] words = sentence.split("\\s+"); for (int i = words.length - 1 ; 0 <= i; i--) { result.append(words[i]).append(' '); } return result.toString().trim(); }
你删除起始空间角色的答案很简单
return reverse.trim();
String.trim()返回字符串的副本，省略前导和尾随空格（从Javadoc文档中复制）。
对于你的整体问题，我做了这个样本：
String job = "This is a job interview question!"; StringBuilder sb = new StringBuilder(job); String[] words = job.split(" "); int i = 0; for (String word : words) { words[i] = (new StringBuilder(word)).reverse().toString(); i++; } System.out.println("job = " + job); System.out.print("rev = "); for (String word: words) { sb.append(new StringBuilder(word).toString()); sb.append(" "); } String rev = sb.toString().trim(); System.out.println(rev);
输出是：
job = This is a job interview question! rev = sihT si a boj weivretni !noitseuq
如果你想更多地包含任何空格字符，例如制表符，换行符，换页符，那么将split()参数更改为split("\\s")因为\s是正则表达式的正则表达式字符类[\ t \ r \ n \ f]。 请注意您必须如何转义正则表达式的Java字符串表示forms中的反斜杠字符（这是split方法所期望的）。
如何在java中反转单词
public class ReverseString { public static void main(String[] args) { String reverse = ""; String original = new String("hidaya"); for ( int i = original.length() - 1 ; i >= 0 ; i-- ) reverse = reverse + original.charAt(i); System.err.println("Orignal string is: "+original); System.out.println("Reverse string is: "+reverse); } }
下面是一种使用流行的split （）函数的编码技术，它可以在所有主要语言中使用，Java toCharArray （），可以完全控制字符串forms的字符，Java StringBuilder类可以提供性能（在C＃中可用）太）。
我认为与其他发布的答案相比，代码更容易理解
public static String reverseWordByWord(String sentence) { StringBuilder result = new StringBuilder(); String[] words = sentence.split("\\s+"); // space(s) are the delimiters for (String word : words) { char[] charArray = word.toCharArray(); int iEnd = word.length() - 1; StringBuilder temp = new StringBuilder(); for (int i = iEnd; i >= 0; i--) { temp.append(charArray[ i]); } result.append(temp); result.append(" "); // separate the words } return result.toString().trim(); // remove the trailing spaces }
提醒作者发布的要求。
样本输入 ：“Hello World”
输出 ：“olleH dlroW”
public String reverseStringWordByWord(String input) { StringBuilder returnValue = new StringBuilder(); int insertIndex = 0; for(int i = 0;i < input.length();i++ ) { if(input.charAt(i)!=' ') { returnValue.insert(insertIndex, currentChar); } else { insertIndex = i+1; returnValue.append(currentChar); } } return returnValue.toString(); }
我自己对Java很陌生，我希望自己被打败了，但我还是觉得我还是试着去试试。 您可以通过构建字符串来解决额外的空白问题，并假设您将在末尾删除不需要的额外空间。 如果考虑性能，那么您可能想重新考虑这个！
编辑：请注意，我的解决方案（现在）处理前导和尾随空格。
public class StringReversal { public static void main(String[] args) { String str = "hello brave new world"; System.out.println("\"" + reverseWordByWord(str) + "\""); } public static String reverseWordByWord(String str) { String reverse = ""; boolean first = true; for (String s : str.split(" ")) { if (first) { first = false; } else { reverse += " "; } StringBuilder sb = new StringBuilder(); for (int i = s.length() - 1; i >= 0; --i) { sb.append(s.charAt(i)); } reverse += sb.toString(); } while (reverse.length() < str.length()) { reverse += " "; } return reverse.substring(0, reverse.length()); } }
所以，我假设你正在学习/练习java，并且家庭作业问题的风险很高……这意味着你要么爱还是恨这个答案……
如果你看一下String对象的源代码，你会在里面找到这样的东西：
private final char value[]; //this stores the String's characters
第一步是获得该值[] ：
char[] myChars = str.toCharArray();
注意函数实现（来自openjdk-7 ），它返回数组的副本而不是原始的副本，因为String对象是不可变的。
public char[] toCharArray() { char result[] = new char[count]; getChars(0, count, result, 0); //Calls System.arraycopy(...) return result; }
现在我们有了myChars我们可以使用它来获得线性时间O（n）的结果！
public static String reverseWordByWord(String str) { char[] myChars = str.toCharArray(); int stringLen = myChars.length; int left = 0, right = 0; for(int index = 0; index < stringLen; index++) { if(chars[index] == ' ') { //assign right reverse(chars, left, right); //update left } } //Don't forget to handle the boundary case (last word in the String)! }
这是反向function：
private static void reverse(char[] chars, int left, int right) { while(left < right) { //Would you know how to swap 2 chars without using a "char tmp" variable? ;) //Update left and right } }
现在只是为了好玩，你可能想要尝试获得以下输出，也许你会从一些有朝一日幻想的面试官那里得到这个确切的问题：
world new brave hello
以下应该在O（n）中进行，而无需任何昂贵的数组复制或重新构造字符数组长度。 处理多个前置，中间和尾随空格。
public class ReverseString { public static void main(String[] args) { String string1 = "hello brave new world"; String string2 = "hello brave new world "; String string3 = " hello brave new world"; String string4 = " hello brave new world "; System.out.println(reverseStringWordByWord(string1)); System.out.println(reverseStringWordByWord(string2)); System.out.println(reverseStringWordByWord(string3)); System.out.println(reverseStringWordByWord(string4)); } private static String reverseStringWordByWord(String string) { StringBuilder sb = new StringBuilder(); int length = string.length(); for(int i=0;i=0; i--) { sb.append(string.charAt(i)); } return sb.toString(); } }
获取String并使用String方法和StringTokenizer对象及其方法，我们可以使用delimeter将String切割成单词。 通过Stack Naturalfunction将所有单词插入(push)到Satck并从Stack中删除(pop)所有单词。 然后打印那些全部。
在这里我们可以采取String s =“你好勇敢的新世界”
import java.util.*; public class StringReverse { public static void main(String[] argv) { String s = "hello brave new world"; Stack myStack = new Stack (); StringTokenizer st = new StringTokenizer(s); while (st.hasMoreTokens()) myStack.push((String) st.nextElement()); // Print the stack backwards System.out.print('"' + s + '"' + " backwards by word is:\n\t\""); while (!myStack.empty()) { System.out.print(myStack.pop()); System.out.print(' '); } System.out.println('"'); } }
如果您使用自己的任何包，请检查上述程序的输出 。
StringBuilder sb = " This is cool"; sb.reverse(); //sb now contains "looc si sihT " System.out.println(sb); for(int i = 0; i < sb.length(); i++) { int index = sb.indexOf(" ", i); // System.out.println(index); if(index > 0) { sb.replace(i, index, new StringBuilder(sb.substring(i, index)).reverse().toString()); i = index; } if(index < 0) { sb.replace(i, sb.length(), new StringBuilder(sb.substring(i, sb.length())).reverse().toString()); break; } } System.out.println(sb); //output "cool is This "
// Create Scanner object Scanner s=new Scanner(System.in); // Take no.of strings that the user wants int n=s.nextInt(); // Create a temp array String temps[]=new String[n]; // Initialize the variable before the user input is stored in it String st=""; // Create a words array String words[]; // Skip first line, if not used user input will be skipped one time s.nextLine(); // Read the no.of strings that user wish to.. for(int k=0;k=0;i--) { // Put each word in user input string from end to start with a space temps[k]+=words[i]+" "; } } // Now print the words! for(int i=0;i
我认为下面的代码比这里提供的任何代码更有效：
public static void revWordsInStringCStyle(String str){ char [] str_ch = str.toCharArray(); System.out.println(str); char temp; int len = str_ch.length; int left = len-1; for(int right =0; right
}
示例：“hello world”
将成为：“dlrow olleho”
这个怎么样 ：
public class Main { public static void main(String args[]){ String input ="***NGuyen**Van******A*******"; String temp = ""; String result =""; for( int i = 0 ; i < input.length() ; i++) { if(input.charAt(i) != '*') { temp = temp + input.charAt(i); } else { if(!temp.equals("")) result = temp + result; result = input.charAt(i) + result ; temp =""; } } System.out.println(result); } } Output: *******A******Van**NGuyen***
我试着没有拆分function。 而是使用子串和for循环。
static String reverseSentenceWithoutSplit(String str){ StringBuilder sb = new StringBuilder(); char [] charArray = str.toCharArray(); int endindex = charArray.length-1; // loop in reverse, char by char for(int i=charArray.length-1; i>=0; i--){ char c = charArray[i]; if(c==' '){ sb.append(str.substring(i + 1, endindex+1)); // substring- start index inclusive, end index exclusive endindex=i-1;// move to first letter sb.append(c); // include the space } if(i==0){ //grab the last word sb.append(str.substring(i, endindex+1)); } } if(sb.length()==0){ // handle case where string has no space return str; } return sb.toString(); }
输入：在你身后是压迫的象征输出：压迫符号a就是你的背后
输入：ThisIsAllOneWord输出：ThisIsAllOneWord
public String reverseEach(String input) { String[] test = input.split(" "); String output=""; for(String t:test) { String p =""; for(int i=t.length()-1;i>=0;i--) { p=p+t.charAt(i); } output=output+p+" "; } return output; }
/* this code uses while loop and the position of spaces come correctly which is a problem if you use for loop */ import java.util.*; class StrWordRev { public void rev(String s) { for(int i=s.length()-1;i>=0;i--) { System.out.print(s.charAt(i)); } System.out.print(" "); } public void main() { Scanner sc=new Scanner(System.in); String s,s1=""; System.out.println("Enter the string : "); s=sc.nextLine(); int i=0; while(i
这应该适合你：
import java.io.*; class internal1 { public static void main(String s[] { DataInputStream dis = new DataInputStream(System.in); try { String a = ""; String b = ""; System.out.print("Enter the string::"); a = dis.readLine(); System.out.print(a.length()); System.out.println(" "); for (int i = 0; i <= a.length() - 1; i++) { if (a.charAt(i) == ' ' || a.charAt(i) == '.') { for (int j = b.length() - 1; j >= 0; j--) { System.out.print(b.charAt(j)); } b = ""; System.out.print(" "); } b = b + a.charAt(i); } } catch (Exception e) { } } }