Search for a Range

Given a sorted array of integers, find the starting and ending position of a given target value.

Your algorithm's runtime complexity must be in the order of O(log n).

If the target is not found in the array, return [-1, -1].

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

思路

通过二分查找，如果没找到返回[-1,-1]如果找到middle，找出start和ending即可

public class Solution {
    public int[] searchRange(int[] A, int target) {
        boolean found = false;
        int low = 0;
        int high = A.length - 1;
        int middle = 0;
        int result[] = new int[2];
        
        while(low <= high){
            middle = (low + high) / 2;
            if(target > A[middle]){
                low = middle + 1;
            }
            else if(target < A[middle]){
                high = middle - 1;
            }
            else
            {
                found = true;
                break;
            }
        }//while
        if(!found)        //没有查找到
        {            
            result[0] = -1;
            result[1] = -1;
        }//if
        else{
            while(middle >= 0 && A[middle] == target){
                middle--;
            }
            if(middle < 0)
                result[0] = 0;
            else
                result[0] = middle + 1;
            middle = result[0];
            while(middle < A.length && A[middle] == target){
                middle++;
            }
            if(middle >= A.length)
                result[1] = A.length - 1;
            else
                result[1] = middle - 1;
        }//else
        return result;
    }
}