我想要反转Java中String的每个单个字（不是整个字符串，只是每个单词）。
示例：如果输入字符串为“Hello World”，则输出应为“olleH dlroW”。
答案 0 :(得分：106)
这应该可以解决问题。这将迭代源字符串中的每个单词，使用StringBuilder的内置reverse()方法将其反转，然后输出反转的单词。
String source = "Hello World";
for (String part : source.split(" ")) {
System.out.print(new StringBuilder(part).reverse().toString());
System.out.print(" ");
}
输出：
olleH dlroW
注意：评论者已经正确地指出了我认为应该在这里提及的一些事情。此示例将在结果的末尾附加一个额外的空格。它还假设您的单词每个单独的空格分隔，并且您的句子不包含标点符号。
答案 1 :(得分：47)
了解您的图书馆; - ）
import org.apache.commons.lang.StringUtils;
String reverseWords(String sentence) {
return StringUtils.reverseDelimited(StringUtils.reverse(sentence), ' ');
}
答案 2 :(得分：23)
您需要在split之后对每个单词执行此操作array个单词。
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);
}
答案 3 :(得分：17)
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自己的作品提交。你要么获得额外的功劳（如果你可以解释它是如何工作的），要么是因为抄袭而被捕（如果你不能）。
答案 4 :(得分：8)
这里没有人考虑使用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单词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"
当您反转\u0308时，使用上述天真方法会将变音字符l移到第一个String之上。您希望它保持在e之上。
答案 5 :(得分：5)
我是一名C / C ++人，在面试时练习java让我知道是否可以更改或改进某些内容。以下内容允许使用多个空格和换行符。
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();
}
这个使用char []。我认为它更有效......
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);
}
答案 6 :(得分：4)
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);
}
答案 7 :(得分：4)
我假设您可以打印结果（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));
}
或者返回反向字符串：
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();
}
答案 8 :(得分：3)
考虑到分隔符可以有多个空格/制表符，并且我们要保留它们：
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();
}
答案 9 :(得分：1)
我在解决问题时想出了这个答案。我尽量不使用嵌套for循环解决方案O（N ^ 2）。我有点强迫自己使用堆栈来获得乐趣：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;
}
答案 10 :(得分：1)
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，你需要将字符串拆分成这样的字样
String sample = "hello world";
String[] words = sample.split(" ");
答案 11 :(得分：0)
您可以使用增强的 for 循环反转字符串，并在里面交换 summands：
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
答案 12 :(得分：0)
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();
}
答案 13 :(得分：0)
这是我的版本，具有相同的数字空格。希望你喜欢它们！
public String reverseWords(String text){
StringBuilder out = new StringBuilder();
for(String word : text.split(" ")){
out.append(new StringBuilder(word).reverse().toString());
out.append(" ");
}
return out.toString().substring(0, out.length() - 1);
}
答案 14 :(得分：0)
我知道这是一个很老的帖子，但是如果能帮到某个人，我想写下答案。请使用以下代码进行正确的解决方案。
public static void main(String[] args) {
// TODO Auto-generated method stub
String string1 = "Hello how are you";
String[] string2 = string1.split(" ");
String result ="";
for(int i=0;i<string2.length;i++)
{
StringBuilder stringBuilder = new StringBuilder(string2[i]);
result = (i==0)?(result+ stringBuilder.reverse().toString()):(result +" "+stringBuilder.reverse().toString());
}
System.out.println("Result line:"+result);
}
控制台中显示的文字： 结果：olleH woh era uoy
答案 15 :(得分：0)
上述某些解决方案具有较高的运行时间复杂性。使用以下算法，可以在O（n）时间内实现。
算法：
复杂性：O（n）其中n是String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);
}
}
答案 16 :(得分：0)
以块为单位反向复制字符串，然后连接空格。 例如。 ＆＃34;你好java世界＆＃34;。
第1块=＆＃34;你好＆＃34;反复复制： - ＆＃34; olleh＆＃34;然后添加空格 第二块=＆＃34; java＆＃34;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。
答案 17 :(得分：0)
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();
}
}
答案 18 :(得分：0)
String input = "Welcome To The Java Programming";
String output = "";
String[] cutAry = input.split("\\s+");
StringBuilder sb = new StringBuilder();
for(String s:cutAry){
sb.append(s);
output += sb.reverse().toString()+" ";
sb.replace(0, sb.length(), "");
}
System.out.println(output);
答案 19 :(得分：0)
具有流API的示例
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;
}
答案 20 :(得分：0)
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();
}
答案 21 :(得分：0)
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String末尾添加的额外空格输出：
This is my sentence
sihT si ym ecnetnes
答案 22 :(得分：0)
package MujeebWorkspace.helps;
// javamujeeb@gmail.com
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]);
}
}
}
}
答案 23 :(得分：0)
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());
}
}
答案 24 :(得分：0)
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);
答案 25 :(得分：0)
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);
执行命令
Love thy neighbor
robhgien yht evoL
答案 26 :(得分：0)
使用split（），您只需要更改要拆分的内容。
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;
}
答案 27 :(得分：0)
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;
}
运行：
olleH dlroW sihT sI s'ynoT edoC
BUILD SUCCESSFUL (total time: 0 seconds)
答案 28 :(得分：-1)
这会反转给定字符串中的单词。假设单词由单个空格分隔。反转是在适当的位置完成的（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);
}
答案 29 :(得分：-2)
如果它反转每个字母而不是这个for循环很好地工作：
for(int i = 0; i < input.length(); i++){
output = output + input.substring((input.length()-i)-1, input.length()-i);
}
答案 30 :(得分：-2)
我希望这对某人有用。
public class ReverseString {
public static void main(String[] args) {
String a="abrakadabra";
String b=reverse(a);
System.out.print(b);
}
public static String reverse(String b){
String j="";
int a=b.length();
for (int x=a;x>0;x--){
j=j+b.substring(x-1,x);}
return j;
}
}
答案 31 :(得分：-2)
简单方法：
String reverseString(String string)
{
String newString = "";
for(int x = string.length() - 1; x > -1; x ++)
newString += string.charAt(x);
return newString;
}
答案 32 :(得分：-2)
最简单的方法之一：
public String reverse(String post)
{
String backward = "";
for(int i = post.length()-1; i >= 0; i--) {
backward = backward + post.substring(i, i+1);
}
return backward;
}