Search for a Range 解答

Question

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].

Solution

Use binary search, first, find left position, then, find right position.

 public class Solution {
     public int[] searchRange(int[] nums, int target) {
         int[] result = new int[2];
         result[0] = -1;
         result[1] = -1;
         if (nums == null || nums.length < 1)
             return result;
         int start = 0, end = nums.length - 1, mid, first, last;
         // Find first position of target
         while (start + 1 < end) {
             mid = (end - start) / 2 + start;
             if (nums[mid] >= target)
                 end = mid;
             else
                 start = mid;
         }
         if (nums[start] == target)
             result[0] = start;
         else if (nums[end] == target)
             result[0] = end;

         // Find last position of target
         start = 0;
         end = nums.length - 1;
         while (start + 1 < end) {
             mid = (end - start) / 2 + start;
             if (nums[mid] <= target)
                 start = mid;
             else
                 end = mid;
         }
         if (nums[end] == target)
             result[1] = end;
         else if (nums[start] == target)
             result[1] = start;
         return result;
     }
 }