Leetcode: Search in rotated sorted array II (search by rotating sorted array 2)

Question:

Follow up for "search in rotated sorted array ":
What ifDuplicatesAre allowed?

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

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

Note:

1) compared with 1, there are only repeated numbers, and binary search is still used as a whole.

2) Method 2:

 

 

 

 

 

Implementation:

I. Implementation:

1 class solution {2 public: 3 bool search (int A [], int N, int target) {4 If (n = 0 | n = 1 & A [0]! = Target) return false; 5 if (a [0] = target) return true; 6 int I = 1; 7 while (A [I-1] <= A [I]) I ++; 8 9 bool pre = binary_search (A, 0, I, target); 10 bool Pos = binary_search (A, I, n-I, target ); 11 return pre = false? POs: Pre; 12} 13 private: 14 bool binary_search (int * B, int Lo, int Len, int goal) 15 {16 int low = lo; 17 int high = lo + len-1; 18 while (low <= high) 19 {20 int middle = (low + high)/2; 21 if (Goal = B [Middle]) // find, return index22 return true; 23 else if (B [Middle] <goal) // on the right side 24 low = middle + 1; 25 else // on the left side 26 high = middle-1; 27} 28 return false; // No, -129} 30 };

2. Open-Source implementation on the Internet:

 1 class Solution { 2 public: 3     bool search(int A[], int n, int target) { 4          int low=0; 5          int high=n-1; 6          while(low<=high) 7          { 8             const int middle=(low+high)/2; 9             if(A[middle]==target) return true;10             if(A[low]<A[middle])11             {12                if(A[low]<=target&&target<A[middle])13                   high=middle-1;14                else15                   low=middle+1;16             }17             else if(A[low]>A[middle])18             {19                if(A[middle]<target&&target<=A[high])20                   low=middle+1;21                else 22                   high=middle-1;23             }24             else25                 low++;26          }27          return false;28     }29 };