61. Search for a Range【medium】

61. Search for a Range【medium】

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

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

Have you met this question in a real interview?

Yes

Example

Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].

Challenge

O(log n) time.

解法一：

 class Solution {
 public:
     /*
      * @param A: an integer sorted array
      * @param target: an integer to be inserted
      * @return: a list of length 2, [index1, index2]
      */
     vector<int> searchRange(vector<int> &A, int target) {
         if (A.empty()) {
             return vector<int>(, -);
         }

         vector<int> result;

         //find first
         int start = ;
         int end = A.size() - ;

         while (start +  < end) {
             int mid = start + (end - start) / ;
             if (A[mid] == target) {
                 end = mid;
             }
             else if (A[mid] < target) {
                 start = mid;
             }
             else if (A[mid] > target) {
                 end = mid;
             }
         }

         if (A[start] == target) {
             result.push_back(start);
         }
         else if (A[end] == target) {
             result.push_back(end);
         }
         else {
             return vector<int>(, -);
         }

         //find last
         start = ;
         end = A.size() - ;

         while (start +  < end) {
             int mid = start + (end - start) / ;
             if (A[mid] == target) {
                 start = mid;
             }
             else if (A[mid] < target) {
                 start = mid;
             }
             else if (A[mid] > target) {
                 end = mid;
             }
         }

         if (A[end] == target) {
             result.push_back(end);
         }
         else if (A[start] == target) {
             result.push_back(start);
         }

         return result;
     }
 };