Leetcode_search-in-rotated-sorted-array-ii(updated c++ and python version)

地址：http://oj.leetcode.com/problems/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.
思路：旋转后的是两个升序的子序列，先找到”断点“，然后根据第一个数，判断在哪一个有序的子数组里做二分查找。
C++ 参考代码：

O(n) running time, second trial is better

class Solution {
public:
bool binary_search(int A[], int n, int target)
{
int left = 0, right = n-1, mid = 0;
while(left<=right)
{
mid = (left+right)/2;
if(target < A[mid])
right = mid-1;
else if(target > A[mid])
left = mid+1;
else
return true;
}
return false;
}
bool search(int A[], int n, int target) {
if(!n)
return false;
if(n==1)
return A[0] == target;
int pos = 0;
while(pos<n-1 && A[pos]<=A[pos+1])
++pos;
if(target > A[0])
return binary_search(A, pos+1, target);
else if(target < A[0])
return binary_search(A+pos+1, n-pos-1, target);
else
return true;
}
};

python参考代码：

class Solution:
def bin_search(self, arr, target):
left = 0
right = len(arr)-1
while left <= right:
mid = (left+right)//2
if target < arr[mid]:
right = mid-1
elif target > arr[mid]:
left = mid+1
else:
return True
return False
# @param A a list of integers
# @param target an integer
# @return a boolean
def search(self, A, target):
if not A:
return False
i = 0
while i<len(A)-1 and A[i]<=A[i+1]:
i+=1
if target > A[0]:
return self.bin_search(A[:i+1], target)
elif target < A[0]:
return self.bin_search(A[i+1:], target)
else:
return True

//SECOND TRIAL, average running time O(lgn), worst time O(n)class Solution {private:    bool rotatedSearch(int A[], int left, int right, int target)    {        if(!A || left > right)            return false;        int mid = (left+right)/2;        if(A[mid]==target || A[left]==target || A[right]==target)            return true;        if(A[left]<A[mid])        {            if(target>A[left] && target<A[mid] && rotatedSearch(A, left+1, mid-1, target))                return true;            return rotatedSearch(A, mid+1, right-1, target);        }        else if(A[left]>A[mid])        {            if(target>A[mid] && target<A[right] && rotatedSearch(A, mid+1, right-1, target))                return true;            return rotatedSearch(A, left+1, mid-1, target);        }        else        {            if(rotatedSearch(A, left+1, mid-1, target))                return true;            return rotatedSearch(A, mid+1, right-1, target);        }    }public:    bool search(int A[], int n, int target) {        return rotatedSearch(A, 0, n-1, target);    }};