Find Minimum in Rotated Sorted Array II

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Follow up for "Find Minimum in Rotated Sorted Array":
What if duplicates are allowed?

Would this affect the run-time complexity? How and why?

Suppose a sorted array 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).

Find the minimum element.

The array may contain duplicates.

分析：

之前Find Minimum in Rotated Sorted Array这篇文章中的分析和解法对于这个问题一样适用，算法复杂度是O(n)。

一样的代码无需任何修改就Accepted了，这个问题的难度级别还被标注为Hard，真是令人费解。

class Solution:

    # @param num, a list of integer
    # @return an integer
    def findMin(self, num):
        pre = num[0]
        for i in num[1:]:
            if i < pre:
                return i
            else:
                pre = i
        return num[0]

if __name__ == ‘__main__‘:
    s = Solution()
    assert s.findMin([5, 5, 6, 6, 0, 1, 2]) == 0
    print ‘PASS‘

小结：

我觉得这两道题目出的有点奇怪，也许出题者的本意有被遗漏的地方？

Find Minimum in Rotated Sorted Array II

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原文地址：http://www.cnblogs.com/openqt/p/4040759.html