如何在int数组中找到最小值的索引？

Question

我对java还是很陌生，所以我想保持简单，我想我必须采用数组的第一个值，然后将其与每个后续值进行比较，如果该值大于第一个值，则替换价值，但我不知道如何从中获取索引。

Answer 1

对于一个非结构化，未排序的数组，假设仅要查找一次最小值，那么您可以做的最好的事情是对所有元素进行一次简单的迭代（ O（n）复杂度），如下所示：

public int findMinIdx(int[] numbers) {
    if (numbers == null || numbers.length == 0) return -1; // Saves time for empty array
    // As pointed out by ZouZou, you can save an iteration by assuming the first index is the smallest
    int minVal = numbers[0] // Keeps a running count of the smallest value so far
    int minIdx = 0; // Will store the index of minVal
    for(int idx=1; idx<numbers.length; idx++) {
        if(numbers[idx] < minVal) {
            minVal = numbers[idx];
            minIdx = idx;
        }
    }
    return minIdx;
}

同样，如果为最小值，则此方法将返回找到的该值的第一种情况的索引。 如果希望它是最后一种情况，只需将numbers[idx] < minVal为numbers[idx] <= minVal 。

Answer 2

这是Java 8

 public static int findMinIdx(int[] numbers) {
        OptionalInt minimun = IntStream.of(numbers).min();
        return   IntStream.of(numbers).boxed().collect(toList()).indexOf(minimun.getAsInt());
    }

Answer 3

从不关心运行时优化，只是在寻找解决方案！，它奏效了，这对您也有帮助，可以找到数组中最低值的索引。

    // array[] -> Received the array in question as an parameter
    // index -> stores the index of the lowest value
    // in for loop, i is important to complete all the comparison in the function
    // When it finds a lower value between the two, it does not change the index
    // When it finds a lower value it changes it's index to that index
    // If array has same value more than once, it will provide index to that values last occurrence
    // Correct me if you find anything not working in this example...
    //...

private static int index_of_minimum_value(int[] array) {
    int index = 0;
    for (int i = 1; i < array.length; i++) {
        if ((array[i - 1] < array[i]) && ([index] > array[i - 1])) index = i - 1;
        else if (array[index] > array[i]) index = i;
    }
    return index;
}