LeetCode 34. Search for a Range Java

题目：

Given an array of integers sorted in ascending order, 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].

题意：给出一个升序排列的数组，其中可能有重复的元素。给出一个target，在数组中查找，若查找到则返回数组中该元素开始的坐标和结束的index(如果其中只有一个元素匹配，则返回的开始和结束index相同,例如[3,3])；若不存在则返回[-1,-1]。我采用的方法是直接对数组进行二分查找，若查找到元素，则对该位置分别向前和向后查找，直到和target不相等；如没有查找到元素则直接返回[-1，-1]. 但是这样做的最差复杂度为0(n)，所以可以使用两个二分查找分别查找最左的target和最右的target，然后相减即可。

代码：

public int[] searchRange(int[] A, int target) {
        int [] res = {-1,-1};
        if(A == null || A.length == 0)
            return res;
        
        //first iteration, find target wherever it is
        int low = 0;
        int high = A.length-1;
        int pos = 0;
        while(low <= high){
            int mid = (low + high)/2;
            pos = mid;
            if(A[mid] > target)
                high = mid - 1;
            else if(A[mid] < target)
                low = mid + 1;
            else{
                res[0] = pos;
                res[1] = pos;
                break;
            }
        }
        
        if(A[pos] != target)
            return res;
        
        //second iteration, find the right boundary of this target
        int newlow = pos;
        int newhigh = A.length-1;
        while(newlow <= newhigh){
            int newmid = (newlow+newhigh)/2;
            if(A[newmid] == target)
                newlow = newmid + 1;
            else
                newhigh = newmid - 1;
        }
        res[1] = newhigh;
        
        //third iteration, find the left boundary of this target
        newlow = 0;
        newhigh = pos;
        while(newlow <= newhigh){
            int newmid = (newlow+newhigh)/2;
            if(A[newmid] == target)
                newhigh = newmid - 1;
            else
                newlow = newmid + 1;
        }
        res[0] = newlow;
        
        return res;
    }