81. Search in Rotated Sorted Array II Leetcode Python

浏览数：67 / 时间：2015年06月08日

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.

generally we can have two methods to do this problem, the first one is based on finding pivot and then divide the array into two parts, then search.

second one is still based on Binary search.

class Solution:
    # @param A a list of integers
    # @param target an integer
    # @return a boolean
    def search(self, A, target):
        if len(A) == 0:
            return False
        low = 0
        high = len(A) - 1
        while low < high:
            mid = low + (high - low) / 2
            if A[mid] == target:
                return True
            if A[low] < A[mid]:
                if A[low] <= target < A[mid]:
                    high = mid - 1
                else:
                    low = mid + 1
            elif A[low] > A[mid]:
                if A[mid]< target <= A[high]:
                    low = mid +1
                else:
                    high = mid -1
            else:
                low += 1
        if A[low] == target:
            return True
        return False

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81. Search in Rotated Sorted Array II Leetcode Python