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 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].

在有序数组中找出指定元素存在的区间。

两次二分，找到等于target和大于target的位置就行了

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        vector<int> result(2,-1);
        int l=lower_bound(nums.begin(),nums.end(),target)-nums.begin();
        int r=upper_bound(nums.begin(),nums.end(),target)-nums.begin();
        if(l!=r)
        {
            result[0]=l;
            result[1]=r-1;
        }
        return result;
    }
};