[转]Number one:单例模式5种JAVA实现及其比较

错误解法一：当处于多线程环境时，会出现竞争。

public class Singleton1     
{
    private Singleton1()
    {
    }
    int a =5;
    private static Singleton1 singleton1 = null;
    public static Singleton1 getSingleton1(){
        if(singleton1 == null){
            singleton1 = new Singleton1();
        }
        return singleton1;
    }
}

错误解法二：虽然加了锁，解决了多线程下的竞争。但是在创建后的每次访问都加锁，影响效率。

public class Singleton2 {
    private Singleton2()
    {
    }
    int a =5;
    private static Singleton2 singleton2 = null;
    public static Singleton2 getSingleton2(){
        synchronized(Singleton2.class)
        {
            if(singleton2 == null){
                singleton2 = new Singleton2();
            }
        }
        return singleton2;
    }
}

可行的解法三：在锁前面再加一个if判断将是一种可行的解法。

public class Singleton3 {
    private Singleton3()
    {
    }
    int a =5;
    private static Singleton3 Singleton3 = null;
    public Singleton3 getSingleton3(){
        if(Singleton3 == null){
            synchronized(Singleton3.class)
            {
                if(Singleton3 == null){
                    Singleton3 = new Singleton3();
                }
            }
        }
        return Singleton3;
    }
}

 推荐的解法4：利用静态构造函数

public class Singleton4 {
    private Singleton4() {
    }

    private static Singleton4 Singleton4 = new Singleton4();

    public Singleton4 getSingleton4() {
        return Singleton4;
    }
}

 推荐的解法5：实现按需创建实例，涉及语言特性

public class Singleton5 {
    Singleton5() {
    }

    public Singleton5 getSingleton5() {
        return Nested.instance;
    }
    
    static class Nested{
        Nested(){
        }
        static Singleton5 instance = new Singleton5();
    }
}