Search for a Range

Posted 2020-06-30

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].

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        int mid,start=0,end=nums.size()-1;        
        vector<int> res;
        while(start<=end){
            mid=start+(end-start)/2;
            if(nums[mid]>target) end=mid-1;
            else if(nums[mid]<target) start=mid+1;
            else{
                int left=mid,right=mid;
                while(left>=0&&nums[left]==target)
                    left--;
                left=left+1;
                while(right<nums.size()&&nums[right]==target)
                    right++;
                right=right-1;
                res.push_back(left);
                res.push_back(right);
                break;
            }
        }
        if(res.empty()){
            res.push_back(-1);
            res.push_back(-1);
        }
        return res;
    }
};