Search for a Range详解编程语言

Given a sorted array of n integers, find the starting and ending position ofa given target value. 
 
If the target is not found in the array, return [-1, -1].

Given [5, 7, 7, 8, 8, 10] and target value 8, return [3, 4]

import java.io.*; 
class test   
{ 
    public static void main (String[] args) throws java.lang.Exception 
    { 
        int arr[] = {5, 7, 7, 8, 8, 10}; 
        int target = 8; 
        int range[] = searchRange(arr, target); 
        System.out.println("range is " + range[0] + " - " + range[1]); 
    } 
    public static int[] searchRange(int[] A, int target) { 
        int[] result = new int[]{-1, -1}; 
        if (A == null || A.length == 0) return result; 
 
        int lb = -1, ub = A.length; 
        // lower bound 
        while (lb + 1 < ub) { 
            int mid = lb + (ub - lb) / 2; 
            if (A[mid] < target) { 
                lb = mid; 
            } else { 
                ub = mid; 
            } 
        } 
        // whether A[lb + 1] == target, check lb + 1 first 
        if ((lb + 1 < A.length) && (A[lb + 1] == target)) { 
            result[0] = lb + 1; 
        } else { 
            result[0] = -1; 
            result[1] = -1; 
            // target is not in the array 
            return result; 
        } 
 
        // upper bound, since ub >= lb, we do not reset lb 
        ub = A.length; 
        while (lb + 1 < ub) { 
            int mid = lb + (ub - lb) / 2; 
            if (A[mid] > target) { 
                ub = mid; 
            } else { 
                lb = mid; 
            } 
        } 
        // target must exist in the array 
        result[1] = ub - 1; 
 
        return result; 
    } 
}

range is 3 - 4