blocks|key|4275680|text|如果你不能洗牌，最简单的方法是使用streams：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4275681|int+count+=+225;
int[]+value+=+new+Random()
++++++++.ints(1,+count+%2B+1)
++++++++.distinct()
++++++++.limit(count)
++++++++.toArray();|code-block|syntax|javascript|4275682|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

If you can't shuffle, the easiest way is to use streams:

<pre><code>int count = 225;
int[] value = new Random()
 .ints(1, count + 1)
 .distinct()
 .limit(count)
 .toArray();
</code></pre>

blocks|key|4275292|text|您可以使用布尔数组来检查添加了哪些值，方法是使用生成的数字作为索引，这将为您提供恒定的查找速度。您只需将数组循环一次。只需使用生成的随机数作为布尔数组的索引，如果该数的索引为false，则我们找到了一个新数。如果生成的数字不包含true，则继续生成另一个随机数，直到找到该数字的false值。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4275293|int+size+=+225;
int+[]+value+=+new+int[size];
boolean+[]+valuesCreated+=+new+boolean[size+%2B+1];

int+randnum;
Random+num+=+new+Random();

for+(int+x+=+0;+x+<+value.length;+x%2B%2B)+//+Fills+array+with+Random+Numbers+from+0+to+225
{
++++do{
++++++++randnum+=+(num.nextInt(value.length))+%2B+1;
++++}while(valuesCreated[randnum]+==+true);

++++value[x]+=+randnum;
++++valuesCreated[randnum]+=+true;
}|code-block|syntax|javascript|4275294|请注意，这会将数组中的值随机放置，但是此方法按顺序生成值的可能性最小，但数组大小越大，这样做的可能性就越小。|4275295|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

You can use a boolean array to check which values have been added by using the number generated as your index which would give you constant lookup speed. You only have to loop the array once. Simply use the random number generated as your index to the boolean array, if the index for that number is false, then we have found a new number. If the number generated does not contain true, then we continue to generate another random number until a false value is found for the number.

<pre><code>int size = 225;
int [] value = new int[size];
boolean [] valuesCreated = new boolean[size + 1];

int randnum;
Random num = new Random();

for (int x = 0; x &lt; value.length; x++) // Fills array with Random Numbers from 0 to 225
{
 do{
 randnum = (num.nextInt(value.length)) + 1;
 }while(valuesCreated[randnum] == true);

 value[x] = randnum;
 valuesCreated[randnum] = true;
}
</code></pre>

Note that this puts the values in the array at random, however there is the smallest of smallest chances that this method will produce the values in order but the larger size array the far less likely it will do so.

blocks|key|4275818|text|您可以在不使用HashMap、Hashset或ArrayList的情况下添加flag并对其进行检查|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4275819|boolean+isThere+=+false;
for+(int+x+=+0;+x+<+value.length;+x%2B%2B)+//+Fills+array+with+Random+Numbers+from+0+to+225
{
++++randnum+=+(num.nextInt(225))+%2B+1;
++++for(int+i+=0;i<value.length;i%2B%2B){
++++++if(value[i]+==+randnum){
++++++++isThere+=+true;
++++++}
++++}
++++if(!isThere){
++++++value[x]+=+randnum;
++++++isThere+=+false;
++++}++++++++
}|code-block|syntax|javascript|4275820|entityMap^0|12|4|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|$]]

You can add <code>flag</code> and check it 
something like that without using HashMap or Hashset or ArrayList

<pre><code>boolean isThere = false;
for (int x = 0; x &lt; value.length; x++) // Fills array with Random Numbers from 0 to 225
{
 randnum = (num.nextInt(225)) + 1;
 for(int i =0;i&lt;value.length;i++){
 if(value[i] == randnum){
 isThere = true;
 }
 }
 if(!isThere){
 value[x] = randnum;
 isThere = false;
 } 
}
</code></pre>

blocks|key|4275848|text|嗯，shmosel是最圆滑的解决方案。但是，如果保持最初尝试的精神，您可以首先使用顺序成员填充数组，然后使用Random实现您自己的shuffle。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|4275849|int+[]+value+=+new+int[225];+
int+randnum;
Random+num+=+new+Random();

for+(int+x+=+0;+x+<+value.length;+x%2B%2B)+
{
++++value[x]+=+x%2B1;
}

for+(int+x+=+0;+x+<+value.length+-+1;+x%2B%2B)
{
++++int+r+=+num.nextInt(225+-+x)+%2B+x;
++++int+tmp+=+value[x];
++++value[x]+=+value[r];
++++value[r]+=+tmp;
}|code-block|syntax|javascript|4275850|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/50633945/315052^0|2|7|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

Well, <a href="https://stackoverflow.com/a/50633945/315052">shmosel</a> is the slickest solution. But, if keeping in the spirit of your original attempt, you can first populate your array with sequential members, and then use Random to implement your own shuffle.

<pre><code>int [] value = new int[225]; 
int randnum;
Random num = new Random();

for (int x = 0; x &lt; value.length; x++) 
{
 value[x] = x+1;
}

for (int x = 0; x &lt; value.length - 1; x++)
{
 int r = num.nextInt(225 - x) + x;
 int tmp = value[x];
 value[x] = value[r];
 value[r] = tmp;
}
</code></pre>

Say I wanted to fill an array with a bunch of random numbers that do not repeat. The requirements are, you must use the Random class from <code>java.util.Random</code>, you cannot use an ArrayList (I already did that using <code>Collections.shuffle</code>. You can only use a single dimensional array and any kind of loop (including if statements). In order to solve this problem, I made an array of 225 and the random numbers were not allowed to exceed 225. Here was the solution I came up with, but it does not seem very efficient. How can I do this quicker? 

I filled the array with random numbers from 1 to 225. I compared each element of the array with every other element and if there was one similarity, I would restart the comparison from the 0 element. I included my source code below. 

<pre><code> int [] value = new int[225]; 
 int randnum;
 Random num = new Random();

 for (int x = 0; x &lt; value.length; x++) // Fills array with Random Numbers from 0 to 225
 {
 randnum = (num.nextInt(225)) + 1;
 value[x] = randnum;
 }

 for (int y = 0; y &lt; value.length; y++) // These two loops compare each value of the array
 {
 for (int z = y + 1; z &lt; value.length; z++)
 {
 while (value[y] == value[z])
 {
 value[y] = num.nextInt(225) + 1;
 y = 0; // If the loop runs, the entire looping process starts over again.
 }
 }
 }
</code></pre>

Fill an array with random non-repeating numbers (How Can I Do This Quicker)?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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假设我想用一堆不重复的随机数填充一个数组。要求是，您必须使用java.util.Random中的Random类，不能使用ArrayList (我已经使用Collections.shuffle做到了。您只能使用一维数组和任何类型的循环(包括if语句)。为了解决这个问题，我做了一个225的数组，随机数不允许超过225。这是...

问用随机的、不重复的数字填充数组(我怎样才能更快地做到这一点)？
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