[BinarySearch]162 Find Peak Elem

• 分类：BinarySearch

• 考察知识点：BinarySearch

• 最优解时间复杂度：O(logn)

• 最优解空间复杂度：O(1)

162. Find Peak Element

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.

代码：

我的方法:

class Solution:
    def findPeakElement(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if len(nums)==0:
            return -1
        
        start=0
        end=len(nums)-1
        while end-start>1:
            mid=(end-start)//2+start
            if nums[mid]>=nums[mid-1] and nums[mid]>=nums[mid+1]:
                return mid
            
            # 说明peak在左边
            if nums[mid-1]>nums[mid]:
                end=mid
            #说明peak在右边
            elif nums[mid+1]>nums[mid]:
                start=mid
                
        if nums[start]>nums[end]:
            return start
        else:
            return end

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讨论：

1.听了这个Geek的思路，虽然印度英语听的我很难受，但是知道这道题用BinarySearch咋解了，然后自己用代码实现！

我真棒棒！