leetcode【162】Find Peak Element

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写在最前面：这可能是我做过的最简单的中等难度的题了

leetcode【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.

没啥好说的

class Solution(object):
    def findPeakElement(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """

        if len(nums) == 1:
            return 0
        for i in range(1,len(nums)-1):
            if nums[i-1] < nums[i] & nums[i] > nums[i+1]:
                return i
        for i in range(0, len(nums) - 1):
            if nums[i] > nums[i+1]:
                return i
        for i in range(0, len(nums)):
            if nums[len(nums)-1] > nums[len(nums)-2]:
                return len(nums)-1

真的很简单，难道有更快的？

我看到有的人用二分，水平可以，就是没必要，找到第一个就可以了好么，要那么复杂干嘛

但是，二分法是很值得讲一讲的

其实我们只需要找到连续的三个数，中间那个比两边大即可。用二分的话，就是从中间开始找第一个峰值。

我们找到一个数，然后向左向右缩小范围，直到left和right两个指针重合。

如果大家理解不了，可以自己手动纸上画一画

二分真的是一种很棒的思想

class Solution(object):
    def findPeakElement(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """

        left  = 0 
        right = len(nums)-1
        while left < right:
            mid = (left + right) // 2
            if nums[mid]  > nums[mid+1]:
                right = mid 
            elif  nums[mid] < nums[mid + 1]:
                left = mid + 1
        return right