Search for a Range C#

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 ofO(logn).

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].

题意：返回数组nums中等于target的元素的开始结束坐标，未找到返回[-1,-1]，要求O(LogN)的时间复杂度。

解析：找到nums中石头有这个元素很简单，因为是sorted的array，折半查找可以找到其中一个元素等于target ，其他的元素也分布在两边。可以使用two pointers，一个向前一个向后，分别找到第一个不等于target的元素。

代码实现：

 public class Solution    {        public int[] SearchRange(int[] nums, int target)        {            if (nums.Length == 0)                return new int[] { -1, -1 };            int begin = 0;            int end = nums.Length - 1;            int mid = (end + begin) / 2;            while (begin < end)            {                if (nums[mid] < target)                {                    begin = mid + 1;                    mid = (begin + end) / 2;                }                else if (nums[mid] > target)                {                    end = mid - 1;                    mid = (begin + end) / 2;                }                else                    break;            }            if (nums[mid] != target)                return new int[] { -1, -1 };            for (begin = mid; begin >= 0; begin--)            {                if (nums[begin] != target)                    break;            }            for (end = mid; end < nums.Length; end++)            {                if (nums[end] != target)                    break;            }            return new int[] { begin + 1, end - 1 };        }    }