反转“Hello World”的每个单词Java 中的字符串

Question

我想反转 Java 中字符串的每个单个单词（不是整个字符串，只是每个单词）。

示例：如果输入字符串是“Hello World”，则输出应为“olleH dlroW”。

Answer 1

简单的方法：

String reverseString(String string)
{
    String newString = "";
    for(int x = string.length() - 1; x > -1; x ++)
        newString += string.charAt(x);
    return newString;
}

Answer 2

这会反转给定字符串中的单词。假定单词之间由一个空格分隔。反转是就地完成的（在字符缓冲区中）。

public static String reversePhrases(String s)
{
    char[] buf = s.toCharArray();
    int len = buf.length;
    int start = 0;
    for (int i = 0; i < len; i++) {
        if (buf[i] == ' ' || i == (len-1)) {
            if (i == (len-1)) {
                i = len;
            }
            int end = (start + i)/2;
            for (int j = start; j < end; j++) {
                char c = buf[j];
                int pos = (start + i) - j - 1;
                buf[j] = buf[pos];
                buf[pos] = c;
            }
            start = i + 1;    
        }
    }
    return new String(buf);
}

Answer 3

String someString = new String("Love thy neighbor");
    System.out.println(someString);
    char[] someChar = someString.toCharArray();
    int j = someChar.length - 1;
    char temp;
    for (int i = 0; i <= someChar.length / 2; i++) {
        temp = someChar[i];
        someChar[i] = someChar[j];
        someChar[j] = temp;
        j--;
    }
    someString = new String(someChar);
    System.out.println(someString);

跑步：

Love thy neighbor
robhgien yht evoL

Answer 4

流 API 示例

    public static String reverseAllWords(String str) {
    String[] arr = str.split(" ");
    String res = "";
    for (int x = 0; x < arr.length; x++) {
        res = res + Stream.of(arr[x].split("")).reduce("", (a, b) -> b + a) + " ";
    }
    return res;
}

Answer 5

您可以使用增强型 for 循环反转字符串并交换其中的加数：

public static String reverse(String str) {
    String revStr = "";
    for (char ch : str.toCharArray()) {
        // concat of chars in reverse order
        revStr = ch + revStr;
    }
    return revStr;
}

public static String[] reverse(String[] arr) {
    String[] revArr = new String[arr.length];
    for (int i = 0; i < arr.length; i++) {
        // same word order
        revArr[i] = reverse(arr[i]);
    }
    return revArr;
}

public static void main(String[] args) {
    String[] arr = reverse("Hello World".split(" "));
    System.out.println(String.join(" ", arr));
}

输出：

olleH dlroW

Answer 6

逐块反向复制字符串，然后连接空格。
例如。 “你好java世界”。

第一个块=“hello”反向复制它：-“olleh”然后添加空格
第二块=“java”等。

public static void main(String args[]) {
    String s, rev = "";
    Scanner in = new Scanner(System.in);

    System.out.println("Enter a string to reverse");
    s = in.nextLine();

    int length = s.length();
    // char[] cs=s.toCharArray();
    int l, r;
    int i = 0;
    while (i < length) {
        l = i; // starting index
        // find length of sub-block to reverse copy
        while (i < length && s.charAt(i) != ' ') { 
            i++;
        }
        r = i - 1; // ending index
        for (int j = r; j >= l; j--) { // copy reverse of sub-block
            rev = rev + s.charAt(j);
        }
        rev = rev + " "; // add the whitespace
        i++;
    }

    System.out.println("Reverse of entered string is: " + rev);
}

程序也适用于单词之间的多个空格。

Answer 7

上述一些解决方案的运行时复杂度较高。使用下面的算法，可以在 O(n) 时间内实现。

算法：

1. 从头到尾解析字符串。
2. 每次遇到空格字符，即“”，将到目前为止解析的字符列表放入一个可以动态增长的ArrayList中。
3. 以相反的顺序打印 ArrayList，这将为您提供所需的输出。

复杂度：O(n)，其中 n 是字符串的长度。

import java.io.IOException;
import java.util.ArrayList;

public class WordReverse {

    public static void main(String[] args) throws IOException {

        String inputStr = "Hello World";
        String reversed = "";
        ArrayList<String> alist = new ArrayList<String>();

        for (int i = inputStr.length() - 1; i >= 0; i--) {
            if (inputStr.charAt(i) != ' ') {
                reversed = reversed + inputStr.charAt(i);
            } else {
                alist.add(reversed);
                reversed = "";
            }
        }
        alist.add(reversed);
        String result = "";

        for (int i = alist.size() - 1; i >= 0; i--) {
            result = result + alist.get(i);
            result = result + " ";
        }
        System.out.println(result);
    }
}

Answer 8

TC - O(n) 和 SC - O(1) 的解决方案

public String reverseString(String str){
    String str = "Hello Wrold";
    int start = 0;

    for (int i = 0; i < str.length(); i++) {

        if (str.charAt(i) == ' ' || i == str.length() - 1) {

            int end = 0;

            if (i == str.length() - 1) {
                end = i;
            } else {
                end = i - 1;
            }

            str = swap(str, start, end);
            start = i + 1;
        }
    }
    System.out.println(str);
   }

 private static String swap(String str, int start, int end) {

    StringBuilder sb = new StringBuilder(str);

    while (start < end) {
        sb.setCharAt(start, str.charAt(end));
        sb.setCharAt(end, str.charAt(start));
        start++;
        end--;
    }
    return sb.toString();
}

Answer 9

 package MujeebWorkspace.helps;
 // [email protected]

 public class Mujeeb {

     static String str= "This code is simple to reverse the word without changing positions";
     static String[] reverse = str.split(" ");

     public static void main(String [] args){  
         reverseMethod();
     }

     public static void reverseMethod(){
         for (int k=0; k<=reverse.length-1; k++) {
             String word =reverse[reverse.length-(reverse.length-k)];
             String subword = (word+" ");
             String [] splitsubword = subword.split("");

             for (int i=subword.length(); i>0; i--){
                 System.out.print(splitsubword[i]);  
             }
         }
     }
 }

Answer 10

with and without api.

public class Reversal {
    public static void main(String s[]){
        String str= "hello world";
        reversal(str);
    }

    static void reversal(String str){
        String s[]=str.split(" ");
        StringBuilder noapi=new StringBuilder();
        StringBuilder api=new StringBuilder();
        for(String r:s){
            noapi.append(reversenoapi(r));
            api.append(reverseapi(r));
        }
        System.out.println(noapi.toString());
        System.out.println(api.toString());
    }

    static String reverseapi(String str){
        StringBuilder sb=new StringBuilder();
        sb.append(new StringBuilder(str).reverse().toString());
        sb.append(' ');
        return sb.toString();

    }

    static String reversenoapi(String str){
        StringBuilder sb=new StringBuilder();
        for(int i=str.length()-1;i>=0;i--){
            sb.append(str.charAt(i));
        }
        sb.append(" ");
        return sb.toString();
    }
}

Answer 11

使用 split() 函数并反转单个单词

    public String reverseSentence(String input)
      {
        String[] words = input.split(" ");
        StringBuilder builder = new StringBuilder();
        for (String s : words)
        {
            String rev = " ";
            for (int i = 0; i < s.length(); i++)
            {
                rev = s.charAt(i) + rev;
            }

            builder.append(rev);
        }

        return builder.toString().trim();
      }

使用 trim() 删除在新字符串末尾添加的额外空格

输出：

    This is my sentence        
    sihT si ym ecnetnes        

Answer 12

public String reverse(String arg) {
    char[] s = arg.toCharArray();
    StringBuilder sb = new StringBuilder();
    boolean reverse = false;
    boolean isChar = false;
    int insertPos = 0;

    for (int i = 0; i < s.length; i++) {
        isChar = Character.isAlphabetic(s[i]);
        if (!reverse && isChar) {
            sb.append(s[i]);
            insertPos = i;
            reverse = true;
        } else if (reverse && isChar) {
            sb.insert(insertPos, s[i]);
        } else if (!reverse && !isChar) {
            sb.append(s[i]);
        } else if (reverse && !isChar) {
            reverse = false;
            sb.append(s[i]);
        }
    }

    return sb.toString();
}

Answer 13

    String input = "Hello World!";

    String temp = "";
    String result = "";

    for (int i = 0; i <= input.length(); i++) {
        if (i != input.length() && input.charAt(i) != ' ') {
            temp = input.charAt(i) + temp;
        } else {
            result = temp + " " + result;
            temp = "";
        }
    }

    System.out.println("the result is: " + result);

Answer 14

class ReverseWordsInString{
    public static String reverse(String s1){
            int l = s1.length();
            if (l>1)
                    return(s1.substring(l-1) + reverse(s1.substring(0,l-1)));
            else
                    return(s1.substring(0));
    }
    public static void main(String[] args){
            String st = "Hello My World!";
            String r = "";
            for (String word : st.split(" "))
                    r += " "+ reverse(word);
            System.out.println("Reversed words in the given string: "+r.trim());
    }
}

Answer 15

使用 split()，您只需更改您想要拆分的内容。

public static String reverseString(String str)
{
    String[] rstr;
    String result = "";
    int count = 0;
    rstr = str.split(" ");
    String words[] = new String[rstr.length];
    for(int i = rstr.length-1; i >= 0; i--)
    {
        words[count] = rstr[i];
        count++;
    }

    for(int j = 0; j <= words.length-1; j++)
    {
        result += words[j] + " ";
    }

    return result;


}

Answer 16

我在解决这个问题时想出了这个答案。我尝试不使用嵌套 for 循环解决方案 O(N^2)。我强迫自己使用堆栈来娱乐：D

    public StringBuilder reverseWord(String input) {
        char separator = ' ';
        char[] chars = input.toCharArray();
        Stack<Character> stack = new Stack<Character>();
        StringBuilder sb = new StringBuilder(chars.length);


        for(int i = 0; i < chars.length; i++) {

            if(chars[i] != separator) { //letters
                stack.push(chars[i]);

                //if not last letter don't go any further
                if(i != chars.length - 1) { continue; }

            }

            while(!stack.isEmpty()) {
                sb.append(stack.pop());
            }
            sb.append(separator);

        }
        //remove the last separator
        sb.deleteCharAt(sb.length() - 1);
        return sb;
    }

Answer 17

public static void main(String[] args) {
        System.out.println(eatWord(new StringBuilder("Hello World This Is Tony's Code"), new StringBuilder(), new StringBuilder()));
    }
static StringBuilder eatWord(StringBuilder feed, StringBuilder swallowed, StringBuilder digested) {
    for (int i = 0, size = feed.length(); i <= size; i++) {
        if (feed.indexOf(" ") == 0 || feed.length() == 0) {
            digested.append(swallowed + " ");
            swallowed = new StringBuilder();
        } else {
            swallowed.insert(0, feed.charAt(0));
        }
        feed = (feed.length() > 0)  ? feed.delete(0, 1) : feed ;
    }
    return digested;
}

跑步：

olleH dlroW sihT sI s'ynoT edoC 
BUILD SUCCESSFUL (total time: 0 seconds)

Answer 18

考虑到分隔符可以是多个空格/制表符，并且我们希望保留它们：

public static String reverse(String string)
{
    StringBuilder sb = new StringBuilder(string.length());
    StringBuilder wsb = new StringBuilder(string.length());
    for (int i = 0; i < string.length(); i++)
    {
        char c = string.charAt(i);
        if (c == '\t' || c == ' ')
        {
            if (wsb.length() > 0)
            {
                sb.append(wsb.reverse().toString());
                wsb = new StringBuilder(string.length() - sb.length());
            }
            sb.append(c);
        }
        else
        {
            wsb.append(c);
        }
    }
    if (wsb.length() > 0)
    {
        sb.append(wsb.reverse().toString());
    }
    return sb.toString();

}

Answer 19

这是一个接受字符串并将其反转的方法。

public String reverse ( String s ) {
            int length = s.length(), last = length - 1;
            char[] chars = s.toCharArray();
            for ( int i = 0; i < length/2; i++ ) {
                char c = chars[i];
                chars[i] = chars[last - i];
                chars[last - i] = c;
            }
            return new String(chars);
        }

首先，您需要将字符串拆分为这样的单词

String sample = "hello world";  
String[] words = sample.split(" ");  

Answer 20

我假设您可以只打印结果（您刚才说“输出应该是...”）;-)

String str = "Hello World";
for (String word : str.split(" "))
    reverse(word);

void reverse(String s) {
    for (int idx = s.length() - 1; idx >= 0; idx--) 
        System.out.println(s.charAt(idx));
}

或者返回反转的字符串：

String str = "Hello World";
StringBuilder reversed = new StringBuilder();
for (String word : str.split(" ")) {
  reversed.append(reverse(word));
  reversed.append(' ');
}
System.out.println(reversed);

String reverse(String s) {
  StringBuilder b = new StringBuilder();
  for (int idx = s.length() - 1; idx >= 0; idx--)
      b.append(s.charAt(idx));
  return b.toString();
}

Answer 21

仅使用 substring() 和递归：

public String rev(String rest) {
    if (rest.equals(""))
        return "";
    return rev(rest.substring(1)) + rest.substring(0,1);
}

Answer 22

好吧，我是一名 C/C++ 人员，在面试中练习 java 让我知道是否可以更改或改进某些内容。以下允许多个空格和换行符。

第一个是使用 StringBuilder

public static String reverse(String str_words){
    StringBuilder sb_result = new StringBuilder(str_words.length());
    StringBuilder sb_tmp = new StringBuilder();
    char c_tmp;
    for(int i = 0; i < str_words.length(); i++){
        c_tmp = str_words.charAt(i);    
        if(c_tmp == ' ' || c_tmp == '\n'){
            if(sb_tmp.length() != 0){   
                sb_tmp.reverse();
                sb_result.append(sb_tmp);
                sb_tmp.setLength(0);
            }   
            sb_result.append(c_tmp);
        }else{
            sb_tmp.append(c_tmp);
        }
    } 
    if(sb_tmp.length() != 0){
        sb_tmp.reverse();
        sb_result.append(sb_tmp);
    }
    return sb_result.toString();
}

这个是使用 char[]。我觉得这样更有效率...

public static String reverse(String str_words){
    char[] c_array = str_words.toCharArray();
    int pos_start = 0;
    int pos_end;
    char c, c_tmp; 
    int i, j, rev_length;
    for(i = 0; i < c_array.length; i++){
        c = c_array[i];
        if( c == ' ' || c == '\n'){
            if(pos_start != i){ 
                pos_end = i-1;
                rev_length = (i-pos_start)/2;
                for(j = 0; j < rev_length; j++){
                    c_tmp = c_array[pos_start+j];
                    c_array[pos_start+j] = c_array[pos_end-j];
                    c_array[pos_end-j] = c_tmp;
                }
            }
            pos_start = i+1;
        }
    }
    //redundant, if only java had '\0' @ end of string
    if(pos_start != i){
        pos_end = i-1;
        rev_length = (i-pos_start)/2;
        for(j = 0; j < rev_length; j++){
            c_tmp = c_array[pos_start+j];
            c_array[pos_start+j] = c_array[pos_end-j];
            c_array[pos_end-j] = c_tmp;
        }
    }   
    return new String(c_array);
}

Answer 23

这里没有人考虑 unicode 字符。您需要使用 java.text.BreakIterator 来查找单词边界，然后在每个单词边界内使用另一个边界来枚举字符边界：

String helloWorld = "He\u0308llo World"; // Hëllo World
StringBuilder reverseStringBuilder = new StringBuilder(helloWorld.length());
BreakIterator wordBreakIterator = BreakIterator.getWordInstance();
wordBreakIterator.setText(helloWorld);

int wordStart = wordIterator.first();
int wordEnd = wordIterator.next();

while (wordEnd != BreakIterator.DONE) {
    String word = helloWorld.substring(wordStart,wordEnd);
    if (Character.isLetterOrDigit(word.charAt(0))) {
        // "Hello" or "World" in our example
        BreakIterator characterBreakIterator = BreakIterator.getCharacterInstance();
        characterBreakIterator.setText(word);
        int characterEnd = characterBreakIterator.last();
        int characterStart = characterBreakIterator.previous();
        while (characterStart != BreakIterator.DONE) {
            reverseStringBuilder.append(word.substring(characterStart, characterEnd));

            characterEnd = characterStart;
            characterStart = characterBreakIterator.previous();
        }
    } else {
        // " " in our example
        reverseStringBuilder.append(word);
    }
    wordStart = wordEnd;
    wordEnd = wordIterator.next();
}

String dlroWolleh = reverseStringBuilder.toString(); // "dlroW ollëH"

使用上面的简单方法将移动变音符号 \u0308String 时，code> 位于第一个 l 之上。您希望它保持在 e 之上。

Answer 24

这是最简单的解决方案，甚至不使用任何循环。

public class olleHdlroW {
    static String reverse(String in, String out) {
        return (in.isEmpty()) ? out :
            (in.charAt(0) == ' ')
            ? out + ' ' + reverse(in.substring(1), "")
            : reverse(in.substring(1), in.charAt(0) + out);
    }
    public static void main(String args[]) {
        System.out.println(reverse("Hello World", ""));
    }
}

即使这是作业，也可以随意复制并作为您自己的作业提交。你要么获得额外的学分（如果你能解释它是如何工作的），要么因抄袭而被抓（如果你不能）。

Answer 25

在分割为单词数组后，您需要对每个单词执行此操作。

public String reverse(String word) {
    char[] chs = word.toCharArray();

    int i=0, j=chs.length-1;
    while (i < j) {
        // swap chs[i] and chs[j]
        char t = chs[i];
        chs[i] = chs[j];
        chs[j] = t;
       i++; j--;
    }
    return String.valueOf(chs);
}

Answer 26

了解您的图书馆;-)

import org.apache.commons.lang.StringUtils;

String reverseWords(String sentence) {
    return StringUtils.reverseDelimited(StringUtils.reverse(sentence), ' ');
}

Answer 27

这应该可以解决问题。这将迭代源字符串中的每个单词，并使用 StringBuilder内置的reverse()方法，并输出反转的单词。

String source = "Hello World";

for (String part : source.split(" ")) {
    System.out.print(new StringBuilder(part).reverse().toString());
    System.out.print(" ");
}

输出：

olleH dlroW 

注释：评论者正确地指出了一些我认为应该在这里提到的事情。此示例将在结果末尾附加一个额外的空格。它还假设您的单词之间用一个空格分隔，并且您的句子不包含标点符号。