Problem: 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.

Solution: Search in Rotated Sorted Array II, Python:

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#!/usr/bin/env python
# -*- encoding: utf-8 -*-
# @Date    : 2015-01-01 22:37:45
# @Author  : NSSimacer
# @Email   : [email protected]
# @Version : 1.0
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
            first = 0
            last = len(A)
            while first != last:
                mid = (first + last) / 2
                if A[mid] == target:
                    return True
                if A[first] < A[mid]:
                    if A[first] <= target and target < A[mid]:
                        last = mid
                    else:
                        first = mid + 1
                elif A[first] > A[mid]:
                    if A[mid] < target and target <= A[last - 1]:
                        first = mid + 1
                    else:
                        last = mid
                else:
                    first += 1  # skip duplicate one
            return False

允许重复元素，在判断的时候，A[first] <= A[mid]，区间[first, mid]的元素不能保证是递增序列，比如[1, 3, 1, 1, 1]。

可以考虑把这个判断条件拆分成两部分：

若A[mid] > A[first]: 区间[first, mid]元素一定递增

若A[mid] == A[first]: 无法判断序列是否递增，将first自增，即向前看一个元素

时间复杂度O(log n)，空间复杂度O(1)。