81. Search in Rotated Sorted Array II

Follow up for "Search in Rotated Sorted Array":
What if duplicates are allowed?

Would this affect the run-time complexity? How and why?

Write a function to determine if a given target is in the array.

 1     bool search(vector<int>& nums, int target)
 2     {
 3         int first = 0;
 4         int last = nums.size();
 5 
 6         while (first != last)
 7         {
 8             const int mid = first + (last - first) / 2;
 9 
10             if (nums[mid] == target)
11             {
12                 return true;
13             }
14 
15             if (nums[first] < nums[mid]) // 此时一定在递增序列中
16             {
17                 if (nums[first] <= target && target < nums[mid])
18                 {
19                     last = mid;
20                 }
21                 else
22                 {
23                     first = mid + 1;
24                 }
25             }
26             else if (nums[first] > nums[mid]) // 此时一定在递增序列中
27             {
28                 if (nums[mid] < target && target <= nums[last - 1])
29                 {
30                     first = mid + 1;
31                 }
32                 else
33                 {
34                     last = mid;
35                 }
36             }
37             else // 不一定在递增序列中 ++后再判断
38             {
39                 ++first;
40             }
41         }
42 
43         return false;
44     }

 

posted @ 2016-07-21 12:19  Ricardo  阅读(87)  评论(0)  编辑  收藏  举报