leetcode: 81. 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?
# Suppose an array sorted in ascending order is rotated 
# at some pivot unknown to you beforehand.
#
# (i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).
#
# Write a function to determine if a given target is in the array.
#
# The array may contain duplicates.

相比 [leetcode: 33. Search in Rotated Sorted Array] ，区别在于：
当 x[mid] = x[left] 时，很难判断 target 会落在哪，那么只能采取 left+=1

class Solution():
    def search(self, x, target):
        left, right = 0, len(x) - 1
        while left <= right:
            mid = (left + right) // 2
            if x[mid] == target:
                return True
            elif x[mid] == x[left]:    # x[mid] = x[left] 时，很难判断 target 会落在哪，那么只能采取 left+=1
                left += 1
            elif x[left] <= target < x[mid] or (x[left] > x[mid] and not x[mid] < target <= x[right]):
                right = mid - 1
            else:
                left = mid + 1
        return False


if __name__ == "__main__":
    assert Solution().search([3, 5, 1], 3) == True
    assert Solution().search([2, 2, 3, 3, 4, 1], 1) == True
    assert Solution().search([4, 4, 5, 6, 7, 0, 1, 2], 5) == True