【leetcode】34. Find the first and last positions of elements in a sorted array

subject ：

34.  Find the first and last positions of the elements in the sort array 
 Given an array of integers in ascending order  nums, And a target value  target. Find the start and end position of the given target value in the array .
 If the target value does not exist in the array  target, return  [-1, -1].
 Advanced ：
 You can design and implement time complexity of  O(log n)  Does the algorithm solve this problem ？

 Example  1：

 Input ：nums = [5,7,7,8,8,10], target = 8
 Output ：[3,4]
 Example  2：

 Input ：nums = [5,7,7,8,8,10], target = 6
 Output ：[-1,-1]
 Example  3：

 Input ：nums = [], target = 0
 Output ：[-1,-1]
 
 Tips ：

0 <= nums.length <= 105
-109 <= nums[i] <= 109
nums  It is a group of non decreasing numbers 
-109 <= target <= 109

Dichotomy ：

class Solution {
    
    public int[] searchRange(int[] nums, int target) {
    
        int n = nums.length;
        if(n == 0) return new int[]{
    -1,-1};
        int l = 0, r = n - 1;
        int res = 0;
        while(l <= r) {
    
            int mid = (l + r) / 2;
            if(target > nums[mid]) l = mid + 1;
            else if(target < nums[mid]) r = mid - 1;
            else {
    
                //  Retain the first equal subscript 
                res = mid;
                //  Traverse to the left 
                while(--mid >= 0 && mid<nums.length && nums[mid] == target) continue;
                //  Traverse to the right 
                while(++res >= 0 && res < n && nums[res] == target) continue;
                return new int[]{
    mid + 1, res - 1};
            }
        }
        return new int[]{
    -1,-1};
    }
}

Time complexity ：O(n)
Spatial complexity ：O(1)