34. Search for a Range

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].

给定一个排好序的数组，找到target出现的开始位置和结束位置并以数组的形式返回。要求时间复杂度为log(n)，所以容易想到用二分查找算法，先找到数组开始的索引，然后再调用二分查找，让target增加1，这个时候调用二分查找函数返回的是经过所有相同的target之后，第一个与target不同的数字，然后用返回的值再减1即可。

int start = SearchRange.BinarySearch(nums,target);    if(nums.length==start||nums[start]!=target){    return new int []{-1,-1};    }        int [] res = new int [2];        res[0] = start;        res[1] = SearchRange.BinarySearch(nums,target+1)-1;        return res;    }    public static int BinarySearch(int [] nums,int target){    int low = 0;        int high = nums.length;        while(low<high){        int mid = low+(high-low)/2;        if(target>nums[mid]){low = mid+1;}        else { high = mid;}        }        return low;