# 產生不重複之亂數
## 解題步驟
### 宣告陣列
* 首先宣告一個陣列空間為10的 num變數
```
int[] num = new int[10];
System.out.print("數字: ");
```
* 透過for 迴圈製作1-10的數字
```
for(int i = 0; i < num.length; i++)
{
num[i] = i + 1;
System.out.print(num[i]+" ");
}
```
### **亂數函式**
* 宣告一個陣列 arr 空間為5
* arr陣列用途為放置亂數
```
int[] arr = new int[5];
```
* 首先使用亂數製作0-9的數字n，並放入num陣列
> 假設亂數產生出 n=2 ，那麼 `num[2] = 3`\
> 以此類推：`num[0]=1, num[1]=2, num[2]=3, …, num[9]=10`
* 再將 `num[2]` 放入 `arr[0]`
| num | \[0] | \[1] | \[2] | \[3] | \[4] | \[5] | \[6] | \[7] | \[8] | \[9] |
| --- | ---- | ---- | ---- | ---- | ---- | ---- | ---- | ---- | ---- | ---- |
| 值 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| arr | \[0] | \[1] | \[2] | \[3] | \[4] |
| --- | ---- | ---- | ---- | ---- | ---- |
| 值 | 3 | 0 | 0 | 0 | 0 |
> 陣列不會因為取代了前方的值，本值就消失\
> 所以最後方\[9]＝10
```
n = (int)(Math.random()*(10-i));
arr[i] = num[n];
```
* 剛剛是透過亂數產生0-9，現在變成產生0-8；目的是不去取用到num\[9]
```
for(int j = n; j < num.length - 1; j++)
{
num[j] = num[j+1];
}
```
> 繼續剛剛的假設，如果產生出 n = 8 那麼 num\[ 8 ]＝10\
> 再把 num\[ 8 ]=10 放到 arr\[1] (此為第二個亂數)。
| num | \[0] | \[1] | \[2] | \[3] | \[4] | \[5] | \[6] | \[7] | \[8] | \[9] |
| --- | ---- | ---- | ---- | ---- | ---- | ---- | ---- | ---- | ---- | ---- |
| 值 | 1 | 2 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 10 |
| arr | \[0] | \[1] | \[2] | \[3] | \[4] |
| --- | ---- | ---- | ---- | ---- | ---- |
| 值 | 3 | 10 | 0 | 0 | 0 |
* 以此類推，再產生0-7、0-6、0-5 …
```
int n;
for(int i = 0; i < arr.length; i++)
{
n = (int)(Math.random()*(10-i));
arr[i] = num[n];
for(int j = n; j < num.length - 1; j++)
{
num[j] = num[j+1];
}
}
return arr;
```
### **完整程式碼**
```
public class Example
{
public static void main(String[] args)
{
int[] num = new int[10];
System.out.print("數字: ");
for(int i = 0; i < num.length; i++)
{
num[i] = i + 1;
System.out.print(num[i]+" ");
}
System.out.print("\n隨機不重複產生5個亂數: ");
int[] Array;
Array = getRandom(num);
for(int i = 0; i < 5; i++)
System.out.print(Array[i] + " ");
}
public static int[] getRandom(int[] num)
{
int[] arr = new int[5];
int n;
for(int i = 0; i < arr.length; i++)
{
n = (int)(Math.random()*(10-i));
arr[i] = num[n];
for(int j = n; j < num.length - 1; j++)
{
num[j] = num[j+1];
}
}
return arr;
}
}
```
### 後記
本文於 iT邦幫忙曾被提出[討論](https://ithelp.ithome.com.tw/questions/10187671)，在此感謝 fysh711426 替我更近一步的解說，以下為 [fysh711426](https://ithelp.ithome.com.tw/users/20106865/profile) 所示範的程式運行：
初值
```
num 1 2 3 4 5 6 7 8 9 10
arr 0 0 0 0 0
```
random n = 5
```
num 1 2 3 4 5 7 8 9 10 10
arr 6 0 0 0 0
```
random n = 5
```
num 1 2 3 4 5 8 9 10 10 10
arr 6 7 0 0 0
```
random n = 3
```
num 1 2 3 5 8 9 10 10 10 10
arr 6 7 4 0 0
```
`10-i` 是因為要讓每次取 num 陣列的範圍變小，\
以上面結果來說明，可以理解成不要取重複 10 的部分。
`num.length – 1` 是因為在把 num 往前移動時，如果不減一，j 在最尾端時 num\[j+1] 就會超出陣列範圍。
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