blocks|key|153645|text|System.currentTimeMillis()算不算外部的？你总是可以得到它并通过一些最大值来计算mod：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|153646|int+rand+=+(int)(System.currentTimeMillis()%25high)%2Blow;|code-block|syntax|javascript|153647|entityMap^0|0|Q|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

Does <code>System.currentTimeMillis()</code> count as external? You could always get this and calculate mod by some max value:

<pre><code>int rand = (int)(System.currentTimeMillis()%high)+low;
</code></pre>

blocks|key|371668|text|你可以使用一个变量的地址，也可以组合多个变量的地址来生成一个更复杂的变量...|type|unstyled|depth|inlineStyleRanges|entityRanges|data|371669|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

You may use the address of a variable or combine the address of more variables to make a more complex one...

blocks|key|371722|text|您可以获得当前的系统时间，但在大多数语言中，这也需要一个函数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|371723|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

You could get the current system time, but that would also require a function in most languages.

blocks|key|153688|text|如果允许您使用一些外部状态(例如，使用当前系统时间进行长时间初始化)，则可以在没有外部函数的情况下完成此操作。这足以让您实现一个简单的伪随机数生成器。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|153689|在每次调用随机函数时，您将使用状态来创建新的随机值，并更新状态，以便后续调用获得不同的结果。|153690|您只需使用常规的Java算术和/或按位操作即可完成此操作，因此不需要任何外部函数。|153691|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|H|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|I|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|J|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|G|$]]

You can do it without external functions if you are allowed to use some external state (e.g. a long initialised with the current system time). This is enough for you to implement a simple psuedo-random number generator.

In each call to your random function, you would use the state to create a new random value, and update the state, so that subsequent calls get different results. 

You can do this with just regular Java arithmetic and/or bitwise operations, so no external functions are required.

blocks|key|153740|text|public+class+randomNumberGenerator+{

++++int+generateRandomNumber(int+min,+int+max)+{
++++++++return+(int)+((System.currentTimeMillis()+%25+max)+%2B+min);
++++}

++++public+static+void+main(String[]+args)+{
++++++++randomNumberGenerator+rn+=+new+randomNumberGenerator();
++++++++int+cv+=+0;
++++++++int+min+=+1,+max+=+4;
++++++++Map<Integer,+Integer>+hmap+=+new+HashMap<Integer,+Integer>();

++++++++int+count+=+min;
++++++++while+(count+<=+max)+{
++++++++++++cv+=+rn.generateRandomNumber(min,+max);
++++++++++++if+((hmap.get(cv)+==+null)+&&+cv+>=+min+&&+cv+<=+max)+{
++++++++++++++++System.out.print(cv+%2B+",");
++++++++++++++++hmap.put(cv,+1);
++++++++++++++++count%2B%2B;
++++++++++++}
++++++++}

++++}
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|153741|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>public class randomNumberGenerator {

 int generateRandomNumber(int min, int max) {
 return (int) ((System.currentTimeMillis() % max) + min);
 }

 public static void main(String[] args) {
 randomNumberGenerator rn = new randomNumberGenerator();
 int cv = 0;
 int min = 1, max = 4;
 Map&lt;Integer, Integer&gt; hmap = new HashMap&lt;Integer, Integer&gt;();

 int count = min;
 while (count &lt;= max) {
 cv = rn.generateRandomNumber(min, max);
 if ((hmap.get(cv) == null) &amp;&amp; cv &gt;= min &amp;&amp; cv &lt;= max) {
 System.out.print(cv + ",");
 hmap.put(cv, 1);
 count++;
 }
 }

 }
}
</code></pre>

blocks|key|153720|text|泊松随机发生器|type|unstyled|depth|inlineStyleRanges|entityRanges|data|153721|假设我们从随机数的期望值'v‘开始。然后，说一个非负整数序列满足具有期望值v的泊松分布，这意味着在子序列上，值的均值(平均值)将出现'v‘。泊松分布是统计的一部分，详细信息可以在维基百科上找到。但是在这里使用这个函数的主要优点是:+1.只生成整数值。2.这些整数的平均值将等于我们最初提供的值。|153722|它在小数值没有意义的应用程序中很有帮助。例如，1分钟内到达机场的飞机数量是2.5(这没有道理)，但这意味着在2分钟内有5个计划到达。|153723|int+poissonRandom(double+expectedValue)+{
++int+n+=+0;+//counter+of+iteration
++double+limit;+
++double+x;++//pseudo+random+number
++limit+=+exp(-expectedValue);
++x+=+rand()+/+INT_MAX;+
++while+(x+>+limit)+{
++++n%2B%2B;
++++x+*=+rand()+/+INT_MAX;
++}
++return+n;
}|code-block|syntax|javascript|153724|这条线|153725|rand()+/+INT_MAX|153726|应该生成一个介于0和1之间的随机数。所以我们可以使用系统的时间。秒/+60将会达到这个目的。我们应该使用哪个函数完全取决于应用程序。|153727|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|U|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|V|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|W|8|@]|9|@]|A|$]]|$1|M|3|N|5|H|7|X|8|@]|9|@]|A|$I|J]]|$1|O|3|P|5|6|7|Y|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

Poisson Random Generator

Lets say we start with an expected value 'v' of the random numbers. Then to say that a sequence of non negative integers satisfies a Poisson Distribution with expected value v means that over subsequences, the mean(average) of the value will appear 'v'.
Poisson Distribution is part of statistics and the details can be found on wikipedia.
But here the main advantage of using this function are:
1. Only integer values are generated.
2. The mean of those integers will be equal to the value we initially provided.

It is helpful in applications where fractional values don't make sense. Like number of planes arriving on an airport in 1min is 2.5(doesn't make sense) but it implies that in 2 mins 5 plans arrive.

<pre><code>int poissonRandom(double expectedValue) {
 int n = 0; //counter of iteration
 double limit; 
 double x; //pseudo random number
 limit = exp(-expectedValue);
 x = rand() / INT_MAX; 
 while (x &gt; limit) {
 n++;
 x *= rand() / INT_MAX;
 }
 return n;
}
</code></pre>

The line

<pre><code>rand() / INT_MAX
</code></pre>

should generate a random number between 0 and 1. So we can use time of the system.
Seconds / 60 will serve the purpose.
Which function we should use is totally application dependent.

This was questions asked in one of the interviews that I recently attended.

As far as I know a random number between two numbers can be generated as follows

<pre><code>public static int rand(int low, int high) {
 return low + (int)(Math.random() * (high - low + 1));
}
</code></pre>

But here I am using Math.random() to generate a random number between 0 and 1 and using that to help me generate between low and high. Is there any other way I can directly do without using external functions?

Generate Random numbers without using any external functions

这是我最近参加的一次面试中提出的问题。据我所知，两个数字之间的随机数可以生成如下public static int rand(int low, int high) {    return low + (int)(Math.random() * (high - low + 1));}但在这里，我使用Math.random()来生成一个介于0和1之间的随机数，并用它来帮助我生成介于低和高之间的值。有没

问无需使用任何外部函数即可生成随机数
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回答 6

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