Leetcode 34. Search for a Range

Search for a Range

• Total accepted:91570
• Total submissions:308037
• Difficulty:medium

Given a sorted array of integers, find the starting and ending position of a Given target value.

Your algorithm ' s runtime complexity must is in the order of O(log n).

If the target is not a found in the array, return [-1, -1] .

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

Idea: two points find the leftmost position of the number that matches the condition, and then traverse backwards to find the interval that satisfies the requirement. Code:

1 classSolution {2  Public:3vector<int> Searchrange (vector<int>& Nums,inttarget) {4         intN=nums.size (), low=0, high=n-1, mid;5vector<int>Res;6          while(low<=High ) {7mid=low+ (high-low)/2;8             if(nums[mid]<target) {9Low=mid+1;Ten             } One             Else{ Ahigh=mid-1; -             } -         } the         if(nums[low]==target) { - res.push_back (low); -              while(Low<n&&nums[low]==target) low++; -Res.push_back (low-1); +         }  -         Else{ +Res.push_back (-1); ARes.push_back (-1); at         } -         returnRes; -     } -};

Leetcode 34. Search for a Range