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 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].

 

 1 class Solution {
 2 public:
 3     vector<int> searchRange(int A[], int n, int target) {
 4         vector<int> range(2, -1);
 5         int lower = 0;
 6         int upper = n;
 7         int mid;
 8 
 9         // Search for lower bound
10         while (lower < upper) {
11             mid = (lower + upper) / 2;
12             if (A[mid] < target)
13                 lower = mid + 1;
14             else
15                 upper = mid;
16         }
17 
18         // If the target is not found, return (-1, -1)
19         if (A[lower] != target)
20             return range;
21         range[0] = lower;
22 
23         // Search for upper bound
24         upper = n;
25         while (lower < upper) {
26             mid = (lower + upper) / 2;
27             if (A[mid] > target)
28                 upper = mid;
29             else
30                 lower = mid + 1;
31         }
32         range[1] = upper - 1;
33 
34         return range;
35     }
36 };

 

posted @ 2013-04-29 16:54  caijinlong  阅读(126)  评论(0)  编辑  收藏  举报