081 Search in Rotated Sorted Array II

在 033  Search in Rotated Sorted Array 稍作改进， 但是最坏情况下 复杂度为O(n)

class Solution:
    # @param A, a list of integers
    # @param target, an integer to be searched
    # @return an integer
    def search(self, A, target):
        if len(A) == 0:
            return False
        return self.solve(A, target, 0, len(A)- 1)

    def solve(self, a, target, start, end):
        if start > end:
            return False
        half = int((end-start)/2) + start
        if a[start] == target:
            return True
        if a[end] == target:
            return True
        if a[half] == target:
            return True
        if a[start] < a[half]:
            if target > a[half] or target < a[start]:
                return self.solve(a, target, half+1, end-1)
            else:
                return self.solve(a, target, start+1, half-1)
        elif a[start] > a[half]:
            if target < a[half] or target > a[start]:
                return self.solve(a, target, start+1, half-1)
            else:
                return self.solve(a, target, half+1, end-1)
        else:
            return self.solve(a, target, start+1, half-1) or self.solve(a, target, half+1, end-1)