"Leetcode" Search for a Range

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 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] .

Due to O (logn) time requirements, it is obvious to find with two points.

The idea is to use a binary search to find one of the target, not found then return [-1,-1]

After finding it, the two-point search is extended from this position to both sides recursively.

Part of the reference to the idea of xiao.he.587.

classSolution { Public: Vector<int> Searchrange (intA[],intNinttarget) {Vector<int> Ret (2, -1); intIND = Helper (A,0, N-1, Target); if(Ind! =-1)        {//Find one target            intleft =IND; intright =IND; ret[0] =Left ; ret[1] =Right ; //recursively extend left towards 0             while(left = Helper (A,0, left-1, target))! =-1) ret[0] =Left ; //recursively extend right towards n-1             while(right = Helper (A, right+1, N-1, target))! =-1) ret[1] =Right ; }        returnret; }    intHelper (intA[],intLowintHighinttarget) {//binary Search         while(Low <=High ) {            intMID = low + (high-low)/2; if(A[mid] = =target)returnmid; Else if(A[mid] >target) high= mid-1; Else Low= mid+1; }        return-1; }};

"Leetcode" Search for a Range