[Leetcode]Find Peak Element

最新推荐文章于 2024-11-13 02:34:54 发布

NoooName

最新推荐文章于 2024-11-13 02:34:54 发布

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本文介绍了一种在数组中寻找峰点元素的有效算法。峰点元素是指大于其左右相邻元素的值。文章首先给出了一种简单但效率较低的线性搜索方法，随后提出了一种基于二分搜索的更高效解决方案，该方案的时间复杂度为O(logn)。

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.

Note:

Your solution should be in logarithmic complexity.

给定一个数组，返回其中一个峰点值~峰点值就是比左右相邻的元素都大的一个元素~最直观的一种解法是线性查找，代码如下，不过时间复杂度是O(n)~

class Solution:
    # @param num, a list of integer
    # @return an integer
    def findPeakElement(self, num):
        if num is None: return None
        for i in xrange(1, len(num)):
            if num[i] < num[i - 1]:
                return i - 1
        return len(num) - 1

题目要求时间复杂度为O(logn)~所以可以用二分搜索来完成~找到一个local maximum

class Solution:
    # @param num, a list of integer
    # @return an integer
    def findPeakElement(self, num):
        if num is None: return None
        l, r = 0, len(num) - 1
        while l < r:
            mid = l + (r - l) / 2
            if num[mid] > num[mid + 1]:
                r = mid
            else:
                l = mid + 1
        return l