Find Peak Element

There is an integer array which has the following features:

• The numbers in adjacent positions are different.
• A[0] < A[1] && A[A.length - 2] > A[A.length - 1].

We define a position P is a peek if:

A[P] > A[P-1] && A[P] > A[P+1]

Find a peak element in this array. Return the index of the peak.

Example

Given [1, 2, 1, 3, 4, 5, 7, 6]

Return index 1 (which is number 2) or 6 (which is number 7)

Note

The array may contains multiple peeks, find any of them.

Challenge

Time complexity O(logN)

Solution

• Time: O(log n)
• 若A[mid] > A[mid - 1] && A[mid] < A[mid + 1]，则找到一个peak为A[mid]；
• 若A[mid - 1] > A[mid]，则A[mid]左侧必定存在一个peak，可用反证法证明：若左侧不存在peak，则A[mid]左侧元素必满足A[0] > A[1] > … > A[mid -1] > A[mid]，与已知A[0] < A[1]矛盾，证毕。
• 同理可得若A[mid + 1] > A[mid]，则A[mid]右侧必定存在一个peak

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class Solution:
    #@param A: An integers list.
    #@return: return any of peek positions.
    def findPeak(self, A):
        # write your code here
        if not A:
            return 0
        front = 0
        end = len(A)-1

        while front <= end:
            mid = (front+end)/2
            if A[mid] > A[mid-1] and A[mid] > A[mid+1]:
                return mid
            elif A[mid] < A[mid-1]:
                end = mid-1
            else:
                front = mid + 1
        return 0