Search for a Range

Desicription

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].

Solution

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class Solution {
public:
vector<int> searchRange(vector<int>& nums, int target) {
vector<int> res;
auto tmp = lower_bound(nums.begin(), nums.end(), target);
if(tmp == nums.end() || *tmp != target){
res.push_back(-1);
res.push_back(-1);
return res;
}
res.push_back(tmp-nums.begin());
tmp = upper_bound(nums.begin(), nums.end(), target);
res.push_back(tmp-nums.begin()-1);
return res;
}
};