最近做了一个活动抽奖需求，项目需要控制预算，概率需要分布均匀，这样才能获得所需要的概率结果。
例如抽奖得到红包奖金，而每个奖金的分布都有一定概率：

现在的问题就是如何根据概率分配给用户一定数量的红包。

一、一般算法

算法思路：生成一个列表，分成几个区间，例如列表长度100，1-40是0.01-1元的区间，41-65是1-2元的区间等，然后随机从100取出一个数，看落在哪个区间，获得红包区间，最后用随机函数在这个红包区间内获得对应红包数。

//per[] = {40,25,20,10,5}
//moneyStr[] = {0.01-1,1-2,2-3,3-5,5-10}
//获取红包金额
public double getMoney(List<String> moneyStr,List<Integer> per){
        double packet = 0.01;
        //获取概率对应的数组下标
        int key = getProbability(per);
        //获取对应的红包值
        String[] moneys = moneyStr.get(key).split("-");

        if (moneys.length < 2){
            return packet;
        }
        
        double min = Double.valueOf(moneys[0]);//红包最小值
        double max = Double.valueOf(moneys[1]);//红包最大值

        Random random = new Random();
        packet = min + (max - min) * random.nextInt(10) * 0.1;

        return packet;
 }

//获得概率对应的key
public int getProbability(List<Integer> per){
        int key = 0;
        if (per == null || per.size() == 0){
            return key;
        }

        //100中随机生成一个数
        Random random = new Random();
        int num = random.nextInt(100);

        int probability = 0;
        int i = 0;
        for (int p : per){
            probability += p;
            //获取落在该区间的对应key
            if (num < probability){
                key = i;
            }
            
            i++;
        }
        
        return key;

    }
    

时间复杂度：预处理O(MN)，随机数生成O(1)，空间复杂度O(MN)，其中N代表红包种类，M则由最低概率决定。

优缺点：该方法优点是实现简单，构造完成之后生成随机类型的时间复杂度就是O(1)，缺点是精度不够高，占用空间大，尤其是在类型很多的时候。

二、离散算法

算法思路：离散算法通过概率分布构造几个点[40, 65, 85, 95,100]，构造的数组的值就是前面概率依次累加的概率之和。在生成1~100的随机数，看它落在哪个区间，比如50在[40,65]之间，就是类型2。在查找时，可以采用线性查找，或效率更高的二分查找。

//per[] = {40, 65, 85, 95,100}
//moneyStr[] = {0.01-1,1-2,2-3,3-5,5-10}
//获取红包金额
public double getMoney(List<String> moneyStr,List<Integer> per){
        double packet = 0.01;
        //获取概率对应的数组下标
        int key = getProbability(per);
        //获取对应的红包值
        String[] moneys = moneyStr.get(key).split("-");

        if (moneys.length < 2){
            return packet;
        }
        
        double min = Double.valueOf(moneys[0]);//红包最小值
        double max = Double.valueOf(moneys[1]);//红包最大值

        Random random = new Random();
        packet = min + (max - min) * random.nextInt(10) * 0.1;

        return packet;
 }

//获得概率对应的key
public int getProbability(List<Integer> per){
        int key = -1;
        if (per == null || per.size() == 0){
            return key;
        }

        //100中随机生成一个数
        Random random = new Random();
        int num = random.nextInt(100);

        int i = 0;
        for (int p : per){
            //获取落在该区间的对应key
            if (num < p){
                key = i;
            }
        }
        
        return key;

    }  

算法复杂度：比一般算法减少占用空间，还可以采用二分法找出R，这样，预处理O(N)，随机数生成O(logN)，空间复杂度O(N)。

优缺点：比一般算法占用空间减少，空间复杂度O(N)。

三、Alias Method

算法思路：Alias Method将每种概率当做一列，该算法最终的结果是要构造拼装出一个每一列合都为1的矩形，若每一列最后都要为1，那么要将所有元素都乘以5（概率类型的数量）。

Alias Method

此时会有概率大于1的和小于1的，接下来就是构造出某种算法用大于1的补足小于1的，使每种概率最后都为1，注意，这里要遵循一个限制：每列至多是两种概率的组合。

最终，我们得到了两个数组，一个是在下面原始的prob数组[0.75,0.25,0.5,0.25,1]，另外就是在上面补充的Alias数组，其值代表填充的那一列的序号索引，（如果这一列上不需填充，那么就是NULL），[4,4,0,1,NULL]。当然，最终的结果可能不止一种，你也可能得到其他结果。

prob[] = [0.75,0.25,0.5,0.25,1]
Alias[] = [4,4,0,1,NULL] (记录非原色的下标)
根据Prob和Alias获取其中一个红包区间。
随机产生一列C，再随机产生一个数R，通过与Prob[C]比较，R较大则返回C，反之返回Alias[C]。

//原概率与红包区间
per[] = {0.25,0.2,0.1,0.05,0.4}
moneyStr[] = {1-2,2-3,3-5,5-10,0.01-1}

举例验证下，比如取第二列，让prob[1]的值与一个随机小数f比较，如果f小于prob[1]，那么结果就是2-3元，否则就是Alias[1]，即4。

我们可以来简单验证一下，比如随机到第二列的概率是0.2，得到第三列下半部分的概率为0.2 * 0.25，记得在第四列还有它的一部分，那里的概率为0.2 * (1-0.25)，两者相加最终的结果还是0.2 * 0.25 + 0.2 * (1-0.25) = 0.2，符合原来第二列的概率per[1]。

import java.util.*;
import java.util.concurrent.atomic.AtomicInteger;

public class AliasMethod {
    /* The random number generator used to sample from the distribution. */
    private final Random random;

    /* The probability and alias tables. */
    private final int[] alias;
    private final double[] probability;

    /**
     * Constructs a new AliasMethod to sample from a discrete distribution and
     * hand back outcomes based on the probability distribution.
     * <p/>
     * Given as input a list of probabilities corresponding to outcomes 0, 1,
     * ..., n - 1, this constructor creates the probability and alias tables
     * needed to efficiently sample from this distribution.
     *
     * @param probabilities The list of probabilities.
     */
    public AliasMethod(List<Double> probabilities) {
        this(probabilities, new Random());
    }

    /**
     * Constructs a new AliasMethod to sample from a discrete distribution and
     * hand back outcomes based on the probability distribution.
     * <p/>
     * Given as input a list of probabilities corresponding to outcomes 0, 1,
     * ..., n - 1, along with the random number generator that should be used
     * as the underlying generator, this constructor creates the probability
     * and alias tables needed to efficiently sample from this distribution.
     *
     * @param probabilities The list of probabilities.
     * @param random        The random number generator
     */
    public AliasMethod(List<Double> probabilities, Random random) {
        /* Begin by doing basic structural checks on the inputs. */
        if (probabilities == null || random == null)
            throw new NullPointerException();
        if (probabilities.size() == 0)
            throw new IllegalArgumentException("Probability vector must be nonempty.");

        /* Allocate space for the probability and alias tables. */
        probability = new double[probabilities.size()];
        alias = new int[probabilities.size()];

        /* Store the underlying generator. */
        this.random = random;

        /* Compute the average probability and cache it for later use. */
        final double average = 1.0 / probabilities.size();

        /* Make a copy of the probabilities list, since we will be making
         * changes to it.
         */
        probabilities = new ArrayList<Double>(probabilities);

        /* Create two stacks to act as worklists as we populate the tables. */
        Stack<Integer> small = new Stack<Integer>();
        Stack<Integer> large = new Stack<Integer>();

        /* Populate the stacks with the input probabilities. */
        for (int i = 0; i < probabilities.size(); ++i) {
            /* If the probability is below the average probability, then we add
             * it to the small list; otherwise we add it to the large list.
             */
            if (probabilities.get(i) >= average)
                large.push(i);
            else
                small.push(i);
        }

        /* As a note: in the mathematical specification of the algorithm, we
         * will always exhaust the small list before the big list.  However,
         * due to floating point inaccuracies, this is not necessarily true.
         * Consequently, this inner loop (which tries to pair small and large
         * elements) will have to check that both lists aren't empty.
         */
        while (!small.isEmpty() && !large.isEmpty()) {
            /* Get the index of the small and the large probabilities. */
            int less = small.pop();
            int more = large.pop();

            /* These probabilities have not yet been scaled up to be such that
             * 1/n is given weight 1.0.  We do this here instead.
             */
            probability[less] = probabilities.get(less) * probabilities.size();
            alias[less] = more;

            /* Decrease the probability of the larger one by the appropriate
             * amount.
             */
            probabilities.set(more,
                    (probabilities.get(more) + probabilities.get(less)) - average);

            /* If the new probability is less than the average, add it into the
             * small list; otherwise add it to the large list.
             */
            if (probabilities.get(more) >= 1.0 / probabilities.size())
                large.add(more);
            else
                small.add(more);
        }

        /* At this point, everything is in one list, which means that the
         * remaining probabilities should all be 1/n.  Based on this, set them
         * appropriately.  Due to numerical issues, we can't be sure which
         * stack will hold the entries, so we empty both.
         */
        while (!small.isEmpty())
            probability[small.pop()] = 1.0;
        while (!large.isEmpty())
            probability[large.pop()] = 1.0;
    }

    /**
     * Samples a value from the underlying distribution.
     *
     * @return A random value sampled from the underlying distribution.
     */
    public int next() {
        /* Generate a fair die roll to determine which column to inspect. */
        int column = random.nextInt(probability.length);

        /* Generate a biased coin toss to determine which option to pick. */
        boolean coinToss = random.nextDouble() < probability[column];

        /* Based on the outcome, return either the column or its alias. */
       /* Log.i("1234","column="+column);
        Log.i("1234","coinToss="+coinToss);
        Log.i("1234","alias[column]="+coinToss);*/
        return coinToss ? column : alias[column];
    }

    public int[] getAlias() {
        return alias;
    }

    public double[] getProbability() {
        return probability;
    }

    public static void main(String[] args) {
        TreeMap<String, Double> map = new TreeMap<String, Double>();

        map.put("1-2", 0.25);
        map.put("2-3", 0.2);
        map.put("3-5", 0.1);
        map.put("5-10", 0.05);
        map.put("0.01-1", 0.4);

        List<Double> list = new ArrayList<Double>(map.values());
        List<String> gifts = new ArrayList<String>(map.keySet());

        AliasMethod method = new AliasMethod(list);
        for (double value : method.getProbability()){
            System.out.println("," + value);
        }

        for (int value : method.getAlias()){
            System.out.println("," + value);
        }

        Map<String, AtomicInteger> resultMap = new HashMap<String, AtomicInteger>();

        for (int i = 0; i < 100000; i++) {
            int index = method.next();
            String key = gifts.get(index);
            if (!resultMap.containsKey(key)) {
                resultMap.put(key, new AtomicInteger());
            }
            resultMap.get(key).incrementAndGet();
        }
        for (String key : resultMap.keySet()) {
            System.out.println(key + "==" + resultMap.get(key));
        }

    }
}

算法复杂度：预处理O(NlogN)，随机数生成O(1)，空间复杂度O(2N)。

优缺点：这种算法初始化较复杂，但生成随机结果的时间复杂度为O(1)，是一种性能非常好的算法。