(java) 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:本题就是一个普通的2分查找的题，找到那个值，再遍历这个位置的前后找到第一个位置和最后一个位置就行

代码如下(已通过leetcode)

public class Solution {
   public int[] searchRange(int[] nums, int target) {
    int[] res;
    int position=genegratelist(nums,target,0,nums.length-1);
    if(position==-1) {
    res=new int[]{-1,-1};
    } else {
    int low=position,high=position;
    while(low>=0&&nums[low]==target) low--;
    while(high<=nums.length-1&&nums[high]==target) high++;
    if(low==high) {
    res=new int[]{position};
    } else res=new int[]{low+1,high-1};
    }
    return res;
   
   }


private int genegratelist(int[] nums, int target, int low,int high) {
// TODO Auto-generated method stub
if(low>high) return -1;
int mid=(low+high)/2;
if(nums[mid]==target) return mid;
else {
if(nums[mid]>target) high=mid-1;
else low=mid+1;
return genegratelist(nums, target, low, high);
}

}
}