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