LeetCode #162: Find Peak Element

本文介绍了一种寻找峰值元素的算法,并提供了两种解决方案:一种是朴素的 O(n) 方法,另一种是利用二分查找优化到 O(log n) 的高效方法。

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Problem Statement

(Source) A peak element is an element that is greater than its neighbours.

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.

Note:
Your solution should be in logarithmic complexity.

Analysis

Naive O(n) solution.

class Solution(object):
    def findPeakElement(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        n = len(nums)
        for i in range(1, n - 1):
            num = nums[i]
            if num > nums[i-1] and num > nums[i+1]:
                return i
        return [0, n - 1][nums[0] < nums[n - 1]]

The optimised O(logn) solution would be using Binary Search.

class Solution(object):
    def findPeakElement(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        return self.search(nums, 0, len(nums) - 1)


    def search(self, nums, start, end):
        if start == end:
            return start
        elif start + 1 == end:
            return [start, end][nums[start] < nums[end]]
        else:
            mid = (start + end) >> 1
            if nums[mid] < nums[mid - 1]:
                return self.search(nums, start, mid - 1)
            elif nums[mid] < nums[mid + 1]:
                return self.search(nums, mid + 1, end)
            else:
                return mid
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