如何在JAVA中随机产生10位不可重复字串的方法！

[wenxichao1]

注意，是10位长度！试过用系统时间跟SessionID截取，但不能保证字串不重复！请高手指点指点！

[qq920999942]

这个问题！！！不是很了解
可以用IF判断不？
或者比较然后再从新输出

[pig_learner]

用时间取毫秒难道还不能达到你的需求？

     org.apache.commons.id.Hex;

	/**
	 * @return String 生成32位的随机数作为id
	 */
	public static String getCode() {
		String s = null;
		try {
			s = new String(Hex.encodeHex(UUID.randomUUID().getRawBytes()));
		} catch (Exception e) {
			e.printStackTrace();
		}
		return s;
	}


  

用这个就永远不会出现相同的ID的。不过是32位的。

[sunnylyy]

import java.util.Random;

public class TestRandom {
    public static Random r = new Random();
        
    public static String getRandom(){
        long num = Math.abs(r.nextLong() % 10000000000L);
        String s = String.valueOf(num);
        for(int i = 0; i < 10-s.length(); i++){
            s = "0" + s;
        }
        
        return s;
    }
    
    public static void main(String[] args) {
        for(int i = 0; i < 10; i++){
            System.out.println(getRandom());
        }
    }
}

结果:
1311368955
8862432610
8165454199
6728821648
0213839374
4422278610
4566304477
6341349613
5823971037
4070959361

[烟雨鹏城]

UUID截取

[wenxichao1]

package pack;

import java.util.HashSet;
import java.util.Random;

public class TestRandom {
public static Random r = new Random();

public static String getRandom(){
long num = Math.abs(r.nextLong() % 10000000000L);
String s = String.valueOf(num);
for(int i = 0; i < 10-s.length(); i++){
s = "0" + s;
}

return s;
}

public static void main(String[] args) {
HashSet set=new HashSet();
for(long i = 0; i < 100000; i++){
String str=getRandom();
set.add(str);

}
for(long i = 0; i < 100000; i++){
String str=getRandom();
set.add(str);

}
for(long i = 0; i < 100000; i++){
String str=getRandom();
set.add(str);

}
for(long i = 0; i < 100000; i++){
String str=getRandom();
set.add(str);

}
System.out.println(set.size());
}
}

结果599983

这个循环次数多了还是会产生重复！还有别的方法没

[xiaowangyao]

有个思路就是说先通过随机函数得到一组符合你想要的字符串.如果其中包含重复字符,那么就任意的冒泡算法把这个重复的字符排除掉.希望能给你点友情提示

[墨水鱼]

要直接生成且不重，只有UUID类。
如果能用Set作为辅助检查是否已经重复，重复了再生成一遍，这样就有很多方法了。
（在apache commons lang里有一个RandomStringUtils可以生成指定长度随机字符串。）

[wenxichao1]

RandomStringUtils 这个产生的字串是乱码，要把它转成什么编码类型？

[closewbq]

我测试了应该没什么重复的。

mport java.util.Random;
public class A {
	public static void main(String args[]){
             for(int i=0;i<100;i++){
            	 System.out.println(GetRandomString(10));
             }
	}

	public static String GetRandomString(int Len) {
		
		String[] baseString={"0","1","2","3",
				"4","5","6","7","8","9",
				"a","b","c","d","e",
				"f","g","h","i","j",
				"k","l","m","n","o",
				"p","q","r","s","t",
				"u","v","w","x","y",
				"z","A","B","C","D",
				"E","F","G","H","I",
				"J","K","L","M","N",
				"O","P","Q","R","S",
				"T","U","V","W","X","Y","Z"};
		Random random = new Random();
	    int length=baseString.length;
        String randomString="";
        for(int i=0;i<length;i++){
        	randomString+=baseString[random.nextInt(length)];
        }
		random = new Random(System.currentTimeMillis());
		String resultStr="";
		for (int i = 0; i < Len; i++) {
			resultStr += randomString.charAt(random.nextInt(randomString.length()-1));
		}
		return resultStr;
	}

}

[墨水鱼]

[Quote=引用 8 楼 wenxichao1 的回复:]
RandomStringUtils 这个产生的字串是乱码，要把它转成什么编码类型？
[/Quote]

RandomStringUtils.random(10, "abcd");
以上面为例，就是产生10位随机字符串，字符串由abcd组成。
即你想要随机字符串包含什么字符，可以写到第二个参数中，不需要什么编码解码

[buqitianxie]

[Quote=引用 9 楼 closewbq 的回复:]
我测试了应该没什么重复的。
Java code
mport java.util.Random;publicclass A {publicstaticvoid main(String args[]){for(int i=0;i<100;i++){
System.out.println(GetRandomString(10));
}
}publicstatic String GetRandomString(int Len) {

String[] baseString={"0","1","2","3","4","5","6","7","8","9","a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z","A","B","C","D","E","F","G","H","I","J","K","L","M","N","O","P","Q","R","S","T","U","V","W","X","Y","Z"};
Random random=new Random();int length=baseString.length;
String randomString="";for(int i=0;i<length;i++){
randomString+=baseString[random.nextInt(length)];
}
random=new Random(System.currentTimeMillis());
String resultStr="";for (int i=0; i< Len; i++) {
resultStr+= randomString.charAt(random.nextInt(randomString.length()-1));
}return resultStr;
}

}
[/Quote]

我在想，如果非要实现完全无重复的字符，可以考虑这个方法：
int lastIndexOf(String str)
在将一个随机产生的新字符添加到字符串上之前，用这个方法校验这个字符是否在现在的串中已经存在，只有在返回值为-1是才执行拼接，这样就可以保证生成的字符串是有完全不同的字符组成。