Search for a Range
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].
题解：很简单，lower_bound()和upper_bound()两个函数用一下就好。
class Solution {
public:
vector<int> searchRange(vector<int>& nums, int target) {
vector<int>ans;
int n=nums.size();
int a=lower_bound(nums.begin(),nums.end(),target)-nums.begin();
if((a==n)||(nums[a]!=target)){
ans.push_back(-);
ans.push_back(-);
return ans;
}
else{
int b=upper_bound(nums.begin(),nums.end(),target)-nums.begin();
ans.push_back(a);
ans.push_back(b-);
return ans;
}
}
};