Leetcode: Find Peak Element

本文介绍了一种在给定数组中寻找峰值元素的算法实现,并确保解决方案的时间复杂度为对数级别。峰值元素定义为大于其邻居的元素,在数组两端设定特殊值以确保峰值的存在。

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Question

A peak element is an element that is greater than its neighbors.

Given an input array where num[i] ≠ num[i+1], find a peak element and return its index.

The array may contain multiple peaks, in that case return the index to any one of the peaks is fine.

You may imagine that num[-1] = num[n] = -∞.

For example, in array [1, 2, 3, 1], 3 is a peak element and your function should return the index number 2.

click to show spoilers.

Note:
Your solution should be in logarithmic complexity.


Analyze:

It is the one encounted in algorithm course before. Have known solution, but still give bugs.

The solution is required to be in logarithmic complexity. It is common to use Divide and conquer method.

Since it assumes num[-1]=num[n] = , there must be one peak in either side if num[mid] is smaller than its neighbor which is in the same side.


Mistake taken

  1. Forget to add shifted index for the second subnum since its goal is to return index of any one peak.
  2. Miss return mid
class Solution:
    # @param nums, an integer[]
    # @return an integer
    def findPeakElement(self, nums):
        return self.help(nums)

    def help(self, subnums):
        if len(subnums)<3:
            return subnums.index(max(subnums))

        mid = len(subnums)/2
        if subnums[mid] < subnums[mid-1]:
            return self.help(subnums[:mid])
        elif subnums[mid] < subnums[mid+1]:
            return (mid+1) + self.help(subnums[mid+1:])
        else:
            return mid

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