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].

public class Solution {    public int[] searchRange(int[] nums, int target) {        //本题是在已排序的数组中进行查找，因此应用二分查找的思想        //牢记二分查找的参数以及函数！        //当找到target数时，进行顺序查找，设置两个指针，一个是left，代表最侧不为target的下标。另一个是right，代表最右侧不为target的下标        //注意int 数组的初始化！int a[]={-1,-1};        int[] result={-1,-1};        if(nums.length<1) return result;        int res=find(nums,target,0,nums.length-1);        if(res==-1){            return result;                    }        int left=res-1;        int right=res+1;        while(left>=0&&nums[left]==target){                 left--;        }         while(right
     
      end) return -1;        int mid=(start+end)/2;        if(nums[mid]==target){            return mid;        }        if(target>nums[mid]){            return find(nums,target,mid+1,end);        }else{            return find(nums,target,start,mid-1);        }    }}