【leetcode】Search in Rotated Sorted Array II

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.

与I类似，只是在二分搜索前，只是在判断左右两边是否有序时更加麻烦一些。

此时复杂度是O(n)了，和采用线性搜索的方法基本一致

 1 class Solution {
 2 public:
 3     bool search(int A[], int n, int target) {
 4        
 5        
 6         int left=0;
 7         int right=n-1;
 8         int mid;
 9         while(left<=right)
10         {
11            
12             mid=(left+right)/2;
13             if(A[mid]==target)
14             {
15                 return true;
16             }
17            
18  
19             //增加了这些判断语句
20             bool isLeft=true;
21            
22             for(int i=left;i<=mid;i++)
23             {
24                 if(A[i]>A[mid])
25                 {
26                     isLeft=false;
27                     break;
28                 }
29             }
30            
31             if(isLeft)
32             {
33                 if(A[left]<=target&&target<A[mid])
34                 {
35                     right=mid-1;
36                 }
37                 else
38                 {
39                     left=mid+1;
40                 }
41             }
42             else
43             {
44                 if(A[mid]<target&&target<=A[right])
45                 {
46                     left=mid+1;
47                 }
48                 else
49                 {
50                     right=mid-1;
51                 }
52             }
53  
54         }
55         return false;
56        
57     }
58 };