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Java冒泡排序算法实现

转载 作者:塔克拉玛干 更新时间:2023-11-03 06:15:54 25 4
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我正在尝试在我的代码中实现冒泡排序算法,以使输出按升序显示。下面是我的代码,后面是我在 JGrasp 中收到的错误。任何建议或指示将不胜感激。

import java.util.*;

public class RandomArray
{
public static void main(String[] args)
{
Scanner input = new Scanner(System.in);
System.out.println("Please enter the number 15 to generate 15 random numbers!");
int randomIn = input.nextInt();
int[] randomNumbers = new int[randomIn];
if(randomIn != 15)
{
System.out.println("The number you entered was not 15. Please run the program again and enter 15...");
}
else if(randomIn == 15)
{
for(int x = 0; x < randomNumbers.length; ++x)
{
randomNumbers[x] = (int) (Math.random()*50);
System.out.println("Your randomly generated numbers are: " + randomNumbers[x]);
}
}

}
public static void bubbleSort(int[] randomNumbers)
{
int n = randomNumbs.length;
int temp = 0;
for(int i = 0; i < n; i++)
{
for(int j = 1; j < (n - 1); j++)
{
if(randomNumbers[j-1] > randomNumbers[j])
{
temp = randomNumbers[j - 1];
randomNumbers[j - 1] = randomNumbers[j];
randomNumbers[j] = temp;
for(int i=0); i<randomNumbers.length; i++)
{
System.out.print(randomNumbers[i] + " ");
}
}
}
}
}


}

我收到的错误是:

----jGRASP exec: javac -g RandomArray.java
RandomArray.java:38: error: ';' expected
for(int i=0); i<randomNumbers.length; i++)
^
RandomArray.java:38: error: not a statement
for(int i=0); i<randomNumbers.length; i++)
^
RandomArray.java:38: error: ')' expected
for(int i=0); i<randomNumbers.length; i++)
^
RandomArray.java:38: error: ';' expected
for(int i=0); i<randomNumbers.length; i++)
^
4 errors

----jGRASP wedge2: exit code for process is 1.
----jGRASP: operation complete.

最佳答案

这就是您实现冒泡排序的方式。我刚刚编写了返回数组的方法。这将按升序排列您的号码:

public int []bubbleSort(int[] arr) {
int size = arr.length;
for (int pass = 1; pass < size; pass++) {
for (int i = 0; i < size-pass; i++) {
if (arr[i] > arr[i+1]) {
int temp = arr[i];
arr[i] = arr[i+1];
arr[i+1] = temp;
}
}

}
return arr;
}

关于Java冒泡排序算法实现,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/41090670/

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