No.34Search for a Range

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]


//my solution

class Solution {
public:
vector<int> searchRange(vector<int>& nums, int target)
{
	vector<int> res;
	if (nums.size() == 0)                                            //若为空，返回【-1，-1】
		return { -1,-1 };
	vector<int>::iterator result = find(nums.begin(), nums.end(), target);        //找到第一个target的位置
	if(result!=nums.end())                                              //如果到不在末尾，插入
	res.push_back(result - nums.begin());
	else if(nums[nums.size() - 1] == target)                      //若末尾为target，插入
		res.push_back(nums.size() - 1);
    else                                                              //否则，返回【-1，-1】
        return {-1,-1};

	if (nums[nums.size() - 1] == target)                              //判断末尾是否为target，是直接插入，返回
	{
		res.push_back(nums.size() - 1);
		return res;
	}
	while (nums[result-nums.begin()] ==target )                       //若不是，往后移动
	{
		result++;
	}
	res.push_back(result - nums.begin()-1);
	return res;

}

};