[LeetCode] Search in Rotated Sorted Array

二分 : 判断条件当a[left] <= a[mid]时，可以肯定a[left..mid]是升序的

循环有序一共有以下两种情况

　　第一种

条件： (A[mid] >= A[low]) ，low~mid 二分，mid~high 递归

第二种

条件： (A[mid] < A[low]) ，low~mid 二分，mid~high 递归

 1 class Solution {
 2 public:
 3     int binarySearch(int A[], int low, int high, int target)
 4     {
 5         int mid = (high+low)/2;
 6         
 7         if(low <= high)
 8         {
 9             if(target == A[mid]) 
10                 return mid;
11             else if(target > A[mid])
12                 return binarySearch(A, mid+1, high, target);
13             else
14                 return binarySearch(A, low, mid-1, target);
15         }
16         return -1;
17     }
18     
19     int search(int A[], int n, int target) 
20     {
21         return search(A, 0, n-1, target);
22     }
23     
24     int search(int A[], int low, int high, int target) 
25     {
26         
27         int mid =  (high+low)/2;
28         
29         if(low > high)
30             return -1;
31         
32         if(A[mid] == target)
33             return mid;
34         
35         if(A[mid] >= A[low]) // the pivot is in the bottom half
36         {
37             if(target >= A[low] && target < A[mid]) // target in the first half
38             {
39                 return binarySearch(A, low, mid-1, target);
40             }
41             else // target in the bottom half
42             {
43                 return search(A, mid+1, high, target);
44             }
45         }
46         else // the pivot is in the first half
47         {
48             if(target >= A[mid+1] && target <= A[high]) // target in the bottom half
49             {
50                 return binarySearch(A, mid+1, high, target);
51             }
52             else // target in the first half
53             {
54                 return search(A, low, mid-1, target);
55             }            
56         }
57     }
58 };

迭代：

判读条件和上面一样

 1 class Solution {
 2 public:
 3     bool search(int A[], int n, int target) {
 4         int low = 0;
 5         int high = n-1;
 6         int mid ;
 7 
 8         
 9         while(low<=high)
10         {
11             mid= (low + high)/2;
12             
13             if(A[mid]==target)
14             {
15                 return true;
16             }
17                     
18             if(A[low]<=A[mid]) // the pivot is in the bottom half
19             {
20                 if(A[low]<=target && target < A[mid]) //target in the first half
21                     high = mid-1;
22                 else
23                     low = mid+1;//target in the bottom half
24             }
25             else // the pivot is in the first half
26             {
27                 if(A[mid] < target && target <= A[high])// the pivot is in the first half
28                     low = mid+1;
29                 else //target in the first half
30                     high = mid-1;
31             }
32             
33         }
34         return false;
35         
36     }
37 };