在没有Arrays.sort的Java中查找数组列表中的最小整数值

Find smallest integer value in array list in Java without Arrays.sort

提问by Giovanni

How can I find the smallest value in a int array without changing the array order?

如何在不更改数组顺序的情况下找到 int 数组中的最小值？

code snippet:

代码片段：

    int[] tenIntArray = new int [10];
    int i, userIn;

    Scanner KyBdIn = new Scanner(System.in);
    System.out.println("Please enter 10 integer numbers ");

    for(i = 0; i < tenIntArray.length; i++){
        System.out.println("Please enter integer " + i);
        userIn = KyBdIn.nextInt();
        tenIntArray[i] = userIn;
    }

I am not sure how I can find the smallest array value in the tenIntArray and display the position

我不确定如何在 tenIntArray 中找到最小的数组值并显示位置

For example the array holds - [50, 8, 2, 3, 1, 9, 8, 7 ,54, 10]

例如数组包含 - [50, 8, 2, 3, 1, 9, 8, 7 ,54, 10]

The output should say "The smallest value is 1 at position 5 in array"

输出应该说“ The smallest value is 1 at position 5 in array”

采纳答案by Alexis C.

Thisfigure should be helpful :

这个数字应该有帮助：

Then to answer your question, what would you do on paper ?

然后回答你的问题，你会在纸上做什么？

1. Create and initialize the min value at tenIntArray[0]
2. Create a variable to hold the index of the min value in the array and initialize it to 0 (because we said in 1. to initialize the min at tenIntArray[0])
3. Loop through the elements of your array
4. If you find an element inferior than the current min, update the minimum value with this element and update the index with the corresponding index of this element
5. You're done

1. 创建并初始化最小值 tenIntArray[0]
2. 创建一个变量来保存数组中最小值的索引并将其初始化为 0（因为我们在 1. 中说过要在 处初始化最小值tenIntArray[0]）
3. 循环遍历数组的元素
4. 如果发现低于当前min的元素，用这个元素更新最小值，用这个元素对应的索引更新索引
5. 你完成了

Writing the algorithm should be straightforward now.

现在编写算法应该很简单了。

回答by user3101937

the first index of a array is zero. not one.

数组的第一个索引为零。不是一个。

for(i = 0; i < tenIntArray.length; i++)

so correct this. the code that you asked is :

所以纠正这一点。你问的代码是：

    int small = Integer.MAX_VALUE;
    int i = 0;
    int index = 0;
    for(int j : tenIntArray){
        if(j < small){
            small = j;
            i++;
            index = i;
        }
    }
    System.out.print("The smallest value is"+small+"at position"+ index +"in array");

回答by user3102604

Try this:

尝试这个：

//Let arr be your array of integers
if (arr.length == 0)
    return;
int small = arr[0];
int index = 0;
for (int i = 0; i < arr.length; i++) {
    if (arr[i] < small) {
        small = arr[i];
        index = i;
    }
}

回答by akhil_mittal

The method I am proposing will find both minand max.

我提议的方法将同时找到min和max。

public static void main(String[] args) {
    findMinMax(new int[] {10,40,50,20,69,37});
}
public static void findMinMax(int[] array) {
    if (array == null || array.length < 1)
        return;
    int min = array[0];
    int max = array[0];

    for (int i = 1; i <= array.length - 1; i++) {
        if (max < array[i]) {
            max = array[i];
        }

        if (min > array[i]) {
            min = array[i];
        }
    }
    System.out.println("min: " + min + "\nmax: " + max);
}

Obviously this is not going to one of the most optimized solution but it will work for you. It uses simple comparison to track minand maxvalues. Output is:

显然，这不是最优化的解决方案之一，但它对您有用。它使用简单的比较来跟踪min和max值。输出是：

min: 10 max: 69

min: 10 max: 69

回答by Dark Army

    int[] input = {12,9,33,14,5,4};

    int max = 0;
    int index = 0;
    int indexOne = 0; 
    int min = input[0];
    for(int i = 0;i<input.length;i++)
    {
        if(max<input[i])
        {
            max = input[i];
            indexOne = i;
        }

        if(min>input[i])
        {
            min = input[i];
            index = i;
        }

    }
    System.out.println(max);
    System.out.println(indexOne);
    System.out.println(min);
    System.out.println(index);

回答by SAM

Here is the function

这是函数

public int getIndexOfMin(ArrayList<Integer> arr){
    int minVal = arr.get(0); // take first as minVal
    int indexOfMin = -1; //returns -1 if all elements are equal
    for (int i = 0; i < arr.size(); i++) {
        //if current is less then minVal
        if(arr.get(i) < minVal ){
            minVal = arr.get(i); // put it in minVal
            indexOfMin = i; // put index of current min
        }
    }
    return indexOfMin;  
}

回答by J.R

Using Java 8Streams you can create a Binary operator which compares two integers and returns smallest among them.

使用Java 8Streams，您可以创建一个二元运算符，它比较两个整数并返回其中最小的一个。

Let arr is your array

让 arr 是你的数组

int[] arr = new int[]{54,234,1,45,14,54};
int small = Arrays.stream(arr).reduce((x, y) -> x < y ? x : y).getAsInt();