[leetcode] 162. Find Peak Element @ python

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原题

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

Given an input array nums, where nums[i] ≠ nums[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 nums[-1] = nums[n] = -∞.

Example 1:

Input: nums = [1,2,3,1]
Output: 2
Explanation: 3 is a peak element and your function should return the index number 2.
Example 2:

Input: nums = [1,2,1,3,5,6,4]
Output: 1 or 5
Explanation: Your function can return either index number 1 where the peak element is 2,
or index number 5 where the peak element is 6.
Note:

Your solution should be in logarithmic complexity.

解法

题意要求logarithmic complexity, 我们想到了二分查找法. 由于左右两边的值都是无穷小, 那么一定存在一个峰值. 先初始化左右两个指针, mid为中间的指针, 如果nums[mid] > nums[mid+1] 大, 那么结果一定在mid的左侧(包含mid), 因此将mid赋值给r; 如果nums[mid] <= nums[mid+1], 那么结果一定在mid的右侧(不包含mid), 因此将mid+1赋值给l. 当 l = r 时, 循环跳出, 我们返回 l
Time: O(log(n))
Space: O(1)

代码

class Solution(object):
    def findPeakElement(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        l, r = 0, len(nums)-1
        while l < r:
            mid = (l+r)//2
            if nums[mid] > nums[mid+1]:
                r = mid
            else:
                l = mid+1
                
        return l