分区数组

Question

给定一个整数数组，找到将数组分区为高数字和低数字的索引。例如[5，-1,3,8,6]，索引3将数组划分为[5，-1,3]和[8,6]，第二个分区中的所有数字都大于第一个。解决方案必须在O（n）中工作。请建议这是否正确并实现O（n）复杂性。

private int calculateWinters(int[] a) {
        System.out.println(partition(a, 0, a.length - 1));
        return partition(a, 0, a.length - 1);

    }

    int partition(int arr[], int low, int high) {
        int pivot = arr[low];
        int i = low; // index of smaller element
        int count = 0;
        for (int j = low; j <= high; j++) {
            // If current element is smaller than or
            // equal to pivot

            if (arr[j] <= pivot) {

                i++;
                i += count;
                count = 0;

            } else {
                int temp = arr[j];
                arr[j] = pivot;
                pivot = temp;
                count++;
            }
        }

        return i;
    }

Answer 1

public static void main(String[] args) {
        Winters w = new Winters();
        int[] a = { 5, 4, 2, 10, 8, 78, 21, 2, 33 };
        w.calculateWinters(a);

    }

    private int calculateWinters(int[] a) {
        System.out.println(partition(a, 0, a.length - 1));
        return partition(a, 0, a.length - 1);

    }

    int partition(int arr[], int low, int high) {
        int pivot = arr[low];
        int i = low; // index of smaller element
        List<Integer> list = new ArrayList<>();
        List<Integer> list1 = new ArrayList<>();
        while (arr[i] <= pivot) {
            list1.add(arr[i]);
            i++;
        }

        for (int k = i; k <= arr.length - 1; k++) {
            list.add(arr[k]);
            if (list1.size() > 1 && arr[k] < Collections.max(list1)) {
                list1.addAll(list);
                i = i + (k - i) + 1;
            }
        }

        return i;
    }

}

Answer 2

    private int calculateWinters(int[] a) {
            System.out.println(partition(a, 0, a.length - 1));
            return partition(a, 0, a.length - 1);

        }

        int partition(int arr[], int low, int high) {
            int pivot = arr[low];
            int i = low; // index of smaller element
            List<Integer> list = new ArrayList<>();// This is the list which will add all the numbers of an array which are smaller than pivot
            List<Integer> list1 = new ArrayList<>();//This will contain all the number which are greater than the pivot
            while (arr[i] <= pivot) {
                list1.add(arr[i]);
                i++;
            }

            for (int k = i; k <= arr.length - 1; k++) {
                list.add(arr[k]);//adding all numbers greater than the pivot.
//if we find any number smaller than the  than the max of already encountered numbers, we will move the index(i) to the current pointer.
                if (list1.size() > 1 && arr[k] < Collections.max(list1)) {
                    list1.addAll(list);
                    i = i + (k - i) + 1;
                }
            }

            return i;
        }

    }