如何找到最大值的数组索引？

The title above sums up my question, to clarify things an example is:

上面的标题总结了我的问题,澄清了一个例子:

array[0] = 1
array[1] = 3
array[2] = 7  // largest
array[3] = 5

so the result I would like is 2, since it contains the largest element 7.

所以我想要的结果是2,因为它包含最大的元素7。

5 个解决方案

#1

public int getIndexOfLargest( int[] array )
{
  if ( array == null || array.length == 0 ) return -1; // null or empty

  int largest = 0;
  for ( int i = 1; i < array.length; i++ )
  {
      if ( array[i] > array[largest] ) largest = i;
  }
  return largest; // position of the first largest found
}

#2

int maxAt = 0;

for (int i = 0; i < array.length; i++) {
    maxAt = array[i] > array[maxAt] ? i : maxAt;
}

#3

public int getIndexOfMax(int array[]) {
    if (array.length == 0) {
        return -1; // array contains no elements
    }
    int max = array[0];
    int pos = 0;

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

#4

one way will be:

一种方式是:

 Integer[] array = new Integer[4];
    array[0] = 1;
    array[1] = 3;
    array[2] = 7;
    array[3] = 5;

    List<Integer> iList = Arrays.asList(array);
    System.out.println(iList.indexOf(Collections.max(iList)));
    System.out.println(iList.indexOf(Collections.min(iList)));

#5

Use Arrays.sort(intArray)

int[] tempArray = myArray.clone();
Arrays.sort(tempArray );
int max = tempArray[tempArray .length - 1];
int result = Arrays.asList(myArray).indexOf(max );