数字在排序数组中出现的次数

题目：

统计一个数字在排序数组中出现的次数。例如：输入排序数组是{1,2,3,3,3,3,4,5}，由于这个3在这个数组中出现了4次，因此输出为4。

解答：

public class Solution {

    public static void main(String[] args) {
        int[] arr1 = {1,2,3,3,3,3,4,5};
        int target1 = 4;

        System.out.println(getNumberOfK(arr1, target1));

        int arr2 = {2,2,2,2,2};
        int target2 = 2;
        System.out.println(getNumberOfK(arr2, target2));
    }

    public static int getNumberOfK(int[] arr, int k) {
        if(arr == null) {
            return 0;
        }

        int number = 0;
        int first = getFirst(arr, k, start, end);
        int lat = getLast(arr, k, start, end);

        if(first > -1 && last > -1) {
            number = last-first+1;
        }

        return number;
    }

    private static int getFirst(int[] arr, int k, int start, int end) {
        if(start > end) {
            return -1;
        }

        int mid = (start+end)>>1;
        if(k == arr[mid]) {
            if(mid > 0 && arr[mid-1] != k) {
                return mid;
            } else {
                end = mid - 1;
            }
        } else if(arr[mid] > k) {
            end = mid - 1;
        } else {
            start = mid + 1;
        }

        return getFirst(arr, k, start, end);
    }
    

    private static int getLast(int[] arr, int k, int start, int end) {
        if(start > end) {
            return -1;
        }

        int mid = (start+end)>>1;
        if(arr[mid] == k) {
            if(mid < arr.length-1 && arr[mid] != k) {
                return mid;
            } else {
                start = mid + 1;
            } else if(arr[mid] > k) {
                end = mid - 1;
            } else {
                start = mid + 1;
            }
        }

        return getLast(arr, k, start, end);
    }
}