LeetCode 34. Search for a Range

Given an array of integers 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].

For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].

直接用lower_bound()来二分大于等于它的位置，用upper_bound()来二分大于它的位置。

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