如何打乱字符串中的字符(例如,hello 可能是 ehlol 或 lleoh 或 ...)。我不想用这个Collections.shuffle(...)
方法,有什么更简单的吗?
问问题
76292 次
16 回答
34
我不知道更简单的事情。但是您可以使用 Math.rand() 功能在字符长度范围内生成一个随机数而无需替换,这会给您一个打乱的输出
public class Shuffle {
public static void main(String[] args) {
Shuffle s = new Shuffle();
s.shuffle("hello");
}
public void shuffle(String input){
List<Character> characters = new ArrayList<Character>();
for(char c:input.toCharArray()){
characters.add(c);
}
StringBuilder output = new StringBuilder(input.length());
while(characters.size()!=0){
int randPicker = (int)(Math.random()*characters.size());
output.append(characters.remove(randPicker));
}
System.out.println(output.toString());
}
}
/*
Sample outputs
hlleo
llheo
leohl
lleho
*/
于 2010-07-23T08:55:06.460 回答
12
不是很好的表现,但在我看来相当可读:
public static String shuffleString(String string)
{
List<String> letters = Arrays.asList(string.split(""));
Collections.shuffle(letters);
String shuffled = "";
for (String letter : letters) {
shuffled += letter;
}
return shuffled;
}
于 2012-06-21T00:52:50.650 回答
10
这个怎么样:
public static String shuffle(String text) {
char[] characters = text.toCharArray();
for (int i = 0; i < characters.length; i++) {
int randomIndex = (int)(Math.random() * characters.length);
char temp = characters[i];
characters[i] = characters[randomIndex];
characters[randomIndex] = temp;
}
return new String(characters);
}
于 2014-05-08T09:01:31.277 回答
6
多么烦人的问题。我终于得到了这个:
import java.util.Collections;
import com.google.common.primitives.Chars;
import org.apache.commons.lang3.StringUtils;
String shuffle(String s) {
List<Character> chars = Chars.asList(s.toCharArray());
Collections.shuffle(chars);
return StringUtils.join(chars.stream().toArray());
}
是的,两个图书馆:)
于 2016-09-03T22:17:15.633 回答
3
例如:
static String shuffle(String text){
if (text.length()<=1)
return text;
int split=text.length()/2;
String temp1=shuffle(text.substring(0,split));
String temp2=shuffle(text.substring(split));
if (Math.random() > 0.5)
return temp1 + temp2;
else
return temp2 + temp1;
}
于 2010-07-23T09:01:38.857 回答
3
class ShuffleString
{
public static String shuffle(String s)
{
String shuffledString = "";
while (s.length() != 0)
{
int index = (int) Math.floor(Math.random() * s.length());
char c = s.charAt(index);
s = s.substring(0,index)+s.substring(index+1);
shuffledString += c;
}
return shuffledString;
}
}
public class foo{
static public void main(String[] args)
{
String test = "hallo";
test = ShuffleString.shuffle(test);
System.out.println(test);
}
}
输出:啊哈
于 2010-07-23T09:09:17.717 回答
2
这是既不需要递归也不需要转换为集合的代码。
public static String shuffle(String string) {
StringBuilder sb = new StringBuilder(string.length());
double rnd;
for (char c: string.toCharArray()) {
rnd = Math.random();
if (rnd < 0.34)
sb.append(c);
else if (rnd < 0.67)
sb.insert(sb.length() / 2, c);
else
sb.insert(0, c);
}
return sb.toString();
}
于 2012-12-14T23:02:53.520 回答
2
不知道为什么你不想使用 shuffle,除非它是为了学校。;)
而且,如果您关心性能,则绝对不能使用任何将字符串与“+”连接的解决方案。
这是我能想到的最紧凑的解决方案:
public static String shuffle(String string) {
if (StringUtils.isBlank(string) {
return string;
}
final List<Character> randomChars = new ArrayList<>();
CollectionUtils.addAll(randomChars, ArrayUtils.toObject(string.toCharArray()));
Collections.shuffle(randomChars);
return StringUtils.join(randomChars, "");
}
于 2013-09-05T21:33:54.577 回答
1
没有外部库,对于那些不介意使用的人Collections.shuffle()
:
static String shuffle(String string){
List<Character> list = string.chars().mapToObj(c -> new Character((char) c))
.collect(Collectors.toList());
Collections.shuffle(list);
StringBuilder sb = new StringBuilder();
list.forEach(c -> sb.append(c));
return sb.toString();
}
于 2018-07-16T14:24:59.443 回答
1
在 Kotlin 中,您可以执行以下操作。
val helloCharArray = "hello".toCharArray()
helloCharArray.shuffle()
val shuffledHello = helloCharArray.concatToString()
于 2020-12-30T17:43:14.303 回答
0
您可以遍历所有字符,将每个字符与下一个字符进行比较。然后如果 Math.rand() > 0.5 将这个字符与下一个交换,否则继续下一个字符。
于 2010-07-23T08:57:40.307 回答
0
String shuffled;
do {
shuffled = Stream.of(text.split("")).sorted((o1, o2) -> ThreadLocalRandom.current().nextInt(3) - 1).collect(Collectors.joining());
}while(shuffled.equals(text));
于 2016-03-13T15:00:34.083 回答
0
String
如果您稍后仍想恢复原件,请尝试以下操作:
public static class ShuffledString
{
private List<Integer> indices;
private String string;
public ShuffledString(List<Integer> indices, String string)
{
this.indices = indices;
this.string = string;
}
public List<Integer> getIndices()
{
return indices;
}
public String getRegularString()
{
StringBuilder stringBuilder = new StringBuilder();
for (int stringIndex = 0; stringIndex < indices.size(); stringIndex++)
{
int characterIndex = indices.indexOf(stringIndex);
stringBuilder.append(string.charAt(characterIndex));
}
return stringBuilder.toString();
}
}
public static ShuffledString shuffle(String input)
{
List<Integer> indices = new ArrayList<>();
StringBuilder output = new StringBuilder(input.length());
while (indices.size() < input.length())
{
int randomIndex;
while (indices.contains(randomIndex = (int) (Math.random() * input.length())))
{
}
indices.add(randomIndex);
output.append(input.charAt(randomIndex));
}
return new ShuffledString(indices, output.toString());
}
于 2017-02-13T19:22:08.373 回答
0
在 Java 8+ 中,这可以通过Collectors.shuffle(...)
以下三行来完成:
代码:
public static String shuffle(final String str) {
List<Character> chars = str.chars().mapToObj(e->(char)e).collect(Collectors.toList());
Collections.shuffle(chars);
return chars.stream().map(e->e.toString()).collect(Collectors.joining());
}
演示:
于 2020-08-01T08:52:22.743 回答
0
Using commons-lang3:
import org.apache.commons.lang3.ArrayUtils;
String shuffle(String text) {
char[] chars = text.toCharArray();
ArrayUtils.shuffle(chars);
return String.valueOf(chars);
}
于 2021-02-09T13:34:25.187 回答
0
One more short implementation using Stream API:
String shuffle(String str) {
return new Random().ints(0, str.length()).distinct().limit(str.length()) // shuffle indexes
.mapToObj(i->""+str.charAt(i)).collect(Collectors.joining()); // collect letters
}
于 2021-05-06T21:31:27.920 回答