[Leetcode] 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 be in the orderO(LogN).

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].

 

Question: according to the time complexity required by the question, use the binary method to set the private two variables begin and end, and record the final range. In the recursive binary process, if the target is found, the table below the target in the array keeps narrowing the begin and expanding the end, so that the begin and end are the maximum range.

The main steps are as follows:

1. If a [Mid] = target, the begin and end values are updated based on the mid value. Then, determine whether the rightmost element of the array on the left is still equal to the target element. If the elements are equal, continue to search for the left array. The array on the right also needs to be processed;
2. If a [Mid] <target, Recursively search the array on the right;
3. If a [Mid]> Target, Recursively search the array on the left;

The search process for the array [,] is as follows:

So the final returned range is [].

The Code is as follows:

 1 public class Solution { 2     private int begin; 3     private int end; 4     public void BinarySearch(int[] A,int target,int s,int e){ 5         if(s > e) 6             return; 7         int mid = s + (e - s)/2; 8         if(target == A[mid]){ 9             if(mid < begin)10                 begin = mid;11             if(mid > end)12                 end = mid;13             if(mid - 1>=0 && A[mid-1] == target)14                 BinarySearch(A, target, s, mid-1);15             if(mid + 1 < A.length && A[mid+1] == target)16                 BinarySearch(A, target, mid+1, e);17         }18         else{19             if(A[mid] > target)20                 BinarySearch(A, target, s, mid-1);21             else {22                 BinarySearch(A, target, mid+1, e);23             }24         }25     }26     public int[] searchRange(int[] A, int target) {27         begin = A.length;28         end = -1;29         BinarySearch(A, target, 0, A.length-1);30         31         int[] answer = new int[2];32         if(begin == A.length && end == -1){33             answer[0] = answer[1] = -1;34         }35         else{36             answer[0] = begin;37             answer[1] = end;38         }39         return answer;40     }41 }