題目：

Given an array of integers nums 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].

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]

程式碼：

 vector<int> searchRange(vector<int>& nums, int target) {
        vector<int> res = { -1,-1 };
	if (nums.size() == 0)return res;
	int i = 0, j = nums.size() - 1,mid=-1;

	if (nums[j]<target || nums[i]>target)return res;
	while (i <= j) {
		mid = (i + j) / 2;
		if (nums[mid] < target) {
			i = mid + 1;
		}
		else if (nums[mid] > target) { 
			j = mid - 1; 
		}
		else {
			while (nums[i] < target)i++;
			while (nums[j] > target)j--;
			res[0] = i; res[1] = j; break;
		}
	}
	return res;
    }
 

注：

先找其中的一個，找到以後左右縮小範圍。

注意考慮特殊情況，當nums.size==0的時候，直接返回[-1,-1]