我如何在獲取元素編號的同時找到數組中的最小編號

Question

我有一個困擾我的問題。 在數組中找到最小的數字很容易，但這是我的問題。 當您在數組中找到最小的數字時，最小的數字將成為數組的第[0]個元素。 我想要的是，我想找到最小的數字，以及它是哪個元素。 這樣的

a[0]   a[1]   a[2]   a[3]
 5      20     1      12

我希望我的代碼寫1 is the smallest number and it's a[2] element of array

我只想從那里得到2 ，所以我可以在其余代碼中使用它。 任何幫助都感激不盡。 感謝大伙們。

編輯：我以前嘗試過這種方式

int[] dizi = {5 , 12, 20, 1};
    int gecici;


    for (int i=0; i<4; i++) {
        for (int y = 1; y<4; y++) {

            if (dizi[i] > dizi[y]) {

                gecici = dizi[i];
                dizi[i] = dizi[y];
                dizi[y] = gecici;

            }

        }

    }

Answer 1

在這種情況下，您可以利用IntStream.range ：

IntStream.range(0, arr.length)
         .mapToObj(index -> new SimpleEntry<>(index, arr[index]))
         .min(Comparator.comparingInt(SimpleEntry::getValue));

例：

int[] arr = new int[]{5,20 ,1 ,12};
IntStream.range(0, arr.length)
         .mapToObj(index -> new SimpleEntry<>(index, arr[index]))
         .min(Comparator.comparingInt(SimpleEntry::getValue))
         .ifPresent(s -> System.out.println(s.getValue()+ 
               " is the smallest number and it's index" + s.getKey() + "of array"));

Answer 2

int[] arr = {3,66,22,44,55};
int small=arr[0];
int index=0;
for(int i=0;i<arr.length;i++){
     if(arr[i]<small){
            small = arr[i];
            index = i;
     }
}

Answer 3

嘗試這個。

int[] a = {5, 20, 1, 12};
IntStream.range(0, a.length)
    .mapToObj(i -> i)
    .min(Comparator.comparing(i -> a[i]))
    .ifPresent(i -> System.out.printf(
        "%d is the smallest number and it's a[%d] element of array%n", a[i], i));

如果您的數組是double，則

double[] a = {5, 20, 1, 12};
IntStream.range(0, a.length)
    .mapToObj(i -> i)
    .min(Comparator.comparing(i -> a[i]))
        .ifPresent(i -> System.out.printf(
            "%f is the smallest number and it's a[%d] element of array%n", a[i], i));

您可以通過一種方法來做到這一點。

/**
 * Returns min index of array x.
 * (Returns -1 when length of array x is zero)
 */
static int findMinIndex(int[] x) {
    return IntStream.range(0, x.length)
        .mapToObj(i -> i)
        .min(Comparator.comparing(i -> x[i]))
        .orElse(-1);
}

像這樣打電話

int[] a = {5, 20, 1, 12};
int minIndex = findMinIndex(a);
System.out.printf("%d is the smallest number and it's a[%d] element of arraay%n",
    a[minIndex], minIndex);

Answer 4

int [] array = {0,1,2,3,4,5,6,7,8,9,10};
int smallestNum=array[0];
int smallestIndex=0;
for(int i=1;i<array[i];i++){
   if(array[i] < smallestNum){ //if you want the last small number then use `<=` (if small number occur multiple times)
      smallestNum = array[i];
      smallestIndex=i;
   }
}

Answer 5

如果要查找從小到大的每個元素及其相應的索引，則可以執行以下操作：

public class Test {
    public static class Pair implements Comparable<Pair> {
        int value;
        int index;
        public Pair(int _value, int _index) {
            this.value = _value;
            this.index = _index;
        }
        @Override
        public int compareTo(Pair that) {
            return Integer.valueOf(this.value).compareTo(Integer.valueOf(that.value));
        }
    }

    public static void main(String args[]) {
        int[] a =new int[]{5, 20, 1, 12};
        int n = a.length;
        Pair[] p = new Pair[n];
        for (int i = 0; i < n; ++i) p[i] = new Pair(a[i], i);
        Arrays.sort(p);
        for (int i = 0; i < n; ++i) {
            System.out.println(i + "th minimum is "+ p[i].value +" and is located at index "+ p[i].index);
        }
    }
}

上述方法的復雜度將是時間復雜度O(n log n) 。 但是，如果您只需要知道最小的索引及其索引，則可以按O(n)時間復雜度輕松地檢索它，如下所示：

   int[] a =new int[]{5, 20, 1, 12};
    int n = a.length;
    int minValue = Integer.MAX_VALUE, minIndex = 0;
    for (int i = 0; i < n; ++i) {
        if (minValue > a[i]) {
            minValue = a[i];
            minIndex = i;
        }
    }
    System.out.println("Minimum value is : "+ minValue+ " and it is located at index: "+ minIndex);