java - 如何从数组中的 search() 方法返回单个打印语句？

Question

我的搜索方法有问题。我想要做的是让我的搜索方法只打印一次语句。因此，如果我的数组不止一次包含“3”，我只想打印“3 was found”。一次而不是检查每个值并报告数组中该点是否存在“3”。我该怎么做？

澄清一下，这就是我所拥有的:

`0,0,0,0,0,0,0,0,0,0
4,9,6,9,0,8,5,2,8,3
Average Value: 5.4
    Maximum Value: 9
    Minimum Value: 0
    3 was not found.
    3 was not found.
    3 was not found.
    3 was not found.
    3 was not found.
    3 was not found.
    3 was not found.
    3 was not found.
    3 was not found.
    3 was found.
    2 was not found.
    2 was not found.
    2 was not found.
    2 was not found.
    2 was not found.
    2 was not found.
    2 was not found.
    2 was found.
    2 was not found.
    2 was not found.`

这就是我想要的:

0,0,0,0,0,0,0,0,0,0

4,9,6,9,0,8,5,2,8,3

Average Value: 5.4
Maximum Value: 9
Minimum Value: 0
3 was found.
2 was not found.

所以这是我的完整类(class)。我创建了一个名为 initialize 的方法，它将为数组中的每个元素分配一个 0 到 10 之间的随机整数；一个叫做 print 的方法打印出我的数组的内容；一个名为 printStats 的方法，用于查找并打印数组中的平均值、最大值和最小值；以及一个名为 search 的方法，它在我的数组中搜索(并打印结果)以查找传递给我的方法的整数参数。

一切正常。

public class ArrayLab
{
    private int[] array;
    public ArrayLab(int numElements)
    {
        array = new int[numElements];
    }
    public void initialize()
    {
       array[0] = (int) (Math.random()*11);
       array[1] = (int) (Math.random()*11);
       array[2] = (int) (Math.random()*11);
       array[3] = (int) (Math.random()*11);
       array[4] = (int) (Math.random()*11);
       array[5] = (int) (Math.random()*11);
       array[6] = (int) (Math.random()*11);
       array[7] = (int) (Math.random()*11);
       array[8] = (int) (Math.random()*11);
       array[9] = (int) (Math.random()*11);
    }
    public void print()  {
        System.out.println(array[0] + "," + array[1] + "," + array[2] + "," + array[3] + "," + array[4] + "," + array[5] + "," + array[6] + "," + array[7] + "," + array[8] + "," + array[9]);
        System.out.println();
   }

       public void printStats()
    {
        double sum = 0;
        int max = 0;
        int min = 0;
        min = array[0];

        for (int i = 0; i < array.length; i++)
        {
            sum = sum + array[i];

            if (array[i] > max)
            {
                max = array[i];
            }
            else if (array[i] < min)
            {
                min = array[i];
            }
        }

        double average = sum/array.length;
        System.out.println("Average Value: " + average);
        System.out.println("Maximum Value: " + max);
        System.out.println("Minimum Value: " + min);


    }

public void search(int numChosen)
{ 
       for(int i = 0; i < array.length; i++)
        {
            if(array[i] == numChosen)
            {
                System.out.println(numChosen + " was found.");
            }
            else
            {
                System.out.println(numChosen + " was not found.");
            }
        }
   }
}

Answer 1

在您第一次成功搜索后，开始使用return 或break 语句来中断循环。

此外，您不应在每次迭代数组时都打印Was Not Found。当您的数组完全耗尽并且找不到搜索查询时，您应该只在最后打印一次。

这是修改后的代码片段:

boolean flag = false;
for(int i = 0; i < array.length; i++)
{
    if(array[i] == numChosen)
    {
        System.out.println(numChosen + " was found.");
        flag = true;
        break;
    }
}

if(!flag) {
    System.out.println(numChosen + " was not found.");
}

或者，您也可以执行以下操作:

for(int i = 0; i < array.length; i++)
{
    if(array[i] == numChosen)
    {
        System.out.println(numChosen + " was found.");
        return;
    }
}
System.out.println(numChosen + " was not found.");