[LeetCode] 162. Find Peak Element_Medium tag: Binary Search

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.

思路就是要么在上升(l = mid), 要么在下降(r = mid), 要么在谷底或者顶点(l = mid or r = mid) 都可以.最后判断A[l] < A[r] , return r else l

04/14/2019 update:

分四种情况：   nums[mid - 1] ? nums[mid]  ? nums[mid + 1]

1)   < > , 峰值，直接返回

2）< <, 上升，取右边，因为nums[-1] = 负无穷

3）> >, 下降，取左边

4）> < , 低谷， 取右边， 可以将2与4合并

Code

class Solution:
    def findPeakElement(self, nums):
        l, r = 0, len(nums)-1  # 可以是0 or len(nums) -1
        while l + 1 < r:
            mid = l + (r-l)//2
            if nums[mid] < nums[mid - 1]:
                r = mid
            elif nums[mid] <  nums[mid + 1]:
                l = mid
            else:
                l = mid  # or r = mid
        if nums[l] < nums[r]:
            return r
        return l

2） update

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