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dubbo的源码地址：

Dubbo中的RandomLoadBalance采用基于权重的随机算法，主要代码如下：

public class RandomLoadBalance extends AbstractLoadBalance {    public static final String NAME = "random";    private final Random random = new Random();    protected 
    
      Invoker
     
       doSelect(List
      
       
        > invokers, URL url, Invocation invocation) {        int length = invokers.size(); // Number of invokers        int totalWeight = 0; // The sum of weights        boolean sameWeight = true; // Every invoker has the same weight?        for (int i = 0; i < length; i++) {            int weight = getWeight(invokers.get(i), invocation);            totalWeight += weight; // Sum            if (sameWeight && i > 0                    && weight != getWeight(invokers.get(i - 1), invocation)) {                sameWeight = false;            }        }        if (totalWeight > 0 && !sameWeight) {            // If (not every invoker has the same weight & at least one invoker's weight>0), select randomly based on totalWeight.            int offset = random.nextInt(totalWeight);            // Return a invoker based on the random value.            for (int i = 0; i < length; i++) {                offset -= getWeight(invokers.get(i), invocation);                if (offset < 0) {                    return invokers.get(i);                }            }        }        // If all invokers have the same weight value or totalWeight=0, return evenly.        return invokers.get(random.nextInt(length));    }}
       
      
     
    

解说：

1）假设有四个集群节点A,B,C,D,对应的权重分别是1,2,3,4,那么请求到A节点的概率就为1/(1+2+3+4) = 10%.B,C,D节点依次类推为20%,30%,40%.

2）虽然这个随机算法理解起来是比较容易的,面试一般不会问这个,但是假如我们要实现类似的功能,他这个代码实现的思路还是很优雅的,非常具有借鉴意义.他这个实现思路从纯数学角度是很好理解的,我们还是按照上面数学分析中的前提条件.我们知道总权重为10(1+2+3+4),那么怎么做到按权重随机呢?根据10随机出一个整数,假如为随机出来的是2.然后依次和权重相减,比如2(随机数)-1(A的权重) = 1,然后1(上一步计算的结果)-2(B的权重) = -1,此时-1 < 0,那么则调用B,其他的以此类推