Description:
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.
Solution:
Brute force, scanning from left to right: Not O(log n) time complexity
Binary search:
inspired by https://www.cnblogs.com/grandyang/p/4217175.html
NOTE: Python2 没有math.inf
class Solution(object):
def findPeakElement(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
length = len(nums)
if length == 1:
return 0
nums = [-float('inf')] + nums + [-float('inf')]
def search(l,r):
if l == r-1:
return l
if l == r-2:
return l + int(nums[l] < nums[l+1])
mid = (l+r)//2
n1,n2,n3 = nums[mid-1],nums[mid],nums[mid+1]
if n2 > n1:
if n2 > n3:
return mid
else:
return search(mid,r)
else:
return search(l,mid)
return search(1,length+1) - 1
Runtime: 24 ms, faster than 97.57% of Python online submissions for Find Peak Element.
Memory Usage: 12.1 MB, less than 6.17% of Python online submissions for Find Peak Element.
