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 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(int A[], int n, int target) {

        // Start typing your C/C++ solution below

        // DO NOT write int main() function

        int min,max,mid;

        int i, j;

        min = 0;

        max = n-1;

        vector<int> result ;

        if(A[0] > target|| A[n-1] <target)

        {

            result.push_back(-1);

            result.push_back(-1);

            return result ;

        }

        while(min <= max)

        {

            mid = min + (max - min)/2;

            if(A[mid] == target)

               break;

             else if(A[mid] < target)

                 min = mid + 1;

               else

                 max = mid - 1;

        }

        

      if( A[mid] == target ) {

            i=j = mid;

            while(i>=0 && A[i] == target)i-- ;

            while(j<= n-1 && A[j] == target) j++ ;

             i++;j--;

        result.push_back(i);

        result.push_back(j) ;

      }else{

       

           result.push_back(-1);

            result.push_back(-1);  

      }

        return result ;

    }

};