数组中最大数和最小数的位置

如何解决数组中最大数和最小数的位置

我不知道如何获得数组中最大和最小数字的索引位置。有人可以帮助我吗？ 我的代码：

int[] array = {4,2,7,6,-3,-1,-2,42,-42,9,-4,5,-5,-6,-7,-8,-99,11,20,1,3};

 int smallest = array[0];
 int largest = array[0];

        for (int i = 1; i < array.length; i++){

            if (array[i] > largest) {

                largest = array[i];

            } else if (array[i] < smallest)

                smallest = array[i];
            
        }
        System.out.println("Largest: " + largest);
        System.out.println("Smallest: " + smallest);

我已经有了最大和最小的数字，但我如何找到索引位置。

解决方法

再创建两个变量来存储最大索引和最小索引，现在每当您在 if-else 语句中为最小和最大分配新值时，也会分配 i 的值。

int smallestInd = 0;
int largestInd = 0;
for (int i = 1; i < array.length; i++){

            if (array[i] > largest) {

                largest = array[i];
                largestInd = i;

            } else if (array[i] < smallest)

                smallest = array[i];
                smallestInd = i;
            
        }

,

这应该有效

    int[] array = {4,2,7,6,-3,-1,-2,42,-42,9,-4,5,-5,-6,-7,-8,-99,11,20,1,3};

    int si,smallest = array[si = 0];
    int li,largest = array[li = 0];

    for (int i = 1; i < array.length; i++) {
        if (array[i] > largest) {
            largest = array[li = i];
        } else if (array[i] < smallest) {
            smallest = array[si = i];
        }
    }
    System.out.println("Largest: " + li);
    System.out.println("Smallest: " + si);

或更容易阅读：

    int[] array = {4,3};

    int si = 0,smallest = array[0];
    int li = 0,largest = array[0];

    for (int i = 1; i < array.length; i++) {
        if (array[i] > largest) {
            largest = array[i];
            li = i;
        } else if (array[i] < smallest) {
            smallest = array[i];
            si = i;
        }
    }
    System.out.println("Largest: " + li);
    System.out.println("Smallest: " + si);

,

这是一个有趣的方法，但它不是很有效，因为它需要对值进行排序。基本上，生成所有可能的索引，然后根据实际数组值对它们进行排序。结果数组的第一个和最后一个元素将分别是最小和最大位置。

int[] array = { 4,3 };

Integer[] arr = IntStream.range(0,array.length).boxed()
        .sorted(Comparator.comparingInt(a -> array[a]))
        .toArray(Integer[]::new);

System.out.println("Index for min = " + arr[0]);
System.out.println("Index for max = " + arr[arr.length-1]);

印刷品

Index for min = 17
Index for max = 18

,

    I was able to came up with two answers : 
    ary.sort();
    console.log(
  `This is max ${ary[0]} of the numbers and here is ${
    ary[-1]
  } the min of the numbers list!`
);
_________________________________________________________________
`   let ary = [8,13,44,55];
    let max = ary[0];
    let min = ary[0];
    let locationMax,locationMin;
    for (i = 0; i <= ary.length; i++) {
      if (ary[i] > max) {
        max = ary[i];
        locationMax = i;
      } else if (ary[i] < min) {
        min = ary[i];
        locationMin = i;
      }
    }
    console.log(
      `This is the location of the max ${locationMax} so The second location number ${locationMin} is for min!`
    );