Array-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].
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class Solution {
    public int[] searchRange(int[] nums, int target) {
        int[] res = new int[]{-1, -1};
        if(nums.length == 0)
            return res;
        
        int start;
        int l = 0;
        int r = nums.length;//二分查找插入位置, r = len;
        int mid = 0;
        while(l < r){
            mid = l + (r - l) / 2;
            if(target > nums[mid]) //二分查找第一个插入位置
                l = mid + 1;
            else
                r = mid;
        }
        if(l == nums.length || nums[l] != target)
            return res;
        start = l;          
        
        int end;
        l++;                        //这一趟从第一个插入位置后面开始查找, 缩小搜索范围
        r = nums.length;
        while(l < r){
            mid = l + (r - l) / 2;
            if(target < nums[mid]) //二分查找最后一个插入位置
                r = mid;
            else
                l = mid + 1;
        }
        end = l - 1;        
        return new int[]{start, end};
    }
}
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