## 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.

## Solutions

这样看条件还挺多。

# Time: O(logn)
# Space: O(1)
class Solution:
def findPeakElement(self, nums: List[int]) -> int:
if not nums:
return -1

n = len(nums)
if n == 1:
return 0
l, r = 0, n - 1
while l < r-1:
mid = l + (r - l) // 2
if nums[mid-1] < nums[mid] > nums[mid+1]:
return mid
if nums[mid+1] > nums[mid]:
l = mid + 1
else:
r = mid - 1
return l if nums[l] > nums[r] else r
# 59/59 cases passed (40 ms)
# Your runtime beats 91.44 % of python3 submissions
# Your memory usage beats 100 % of python3 submissions (12.7 MB)


有大牛的简洁版：

# Time: O(logn)
# Space: O(1)
class Solution:
def findPeakElement(self, nums: List[int]) -> int:
if not nums:
return -1
n = len(nums)
if n == 1:
return 0
l, r = 0, n - 1
while l < r:
mid = l + (r - l) // 2
if nums[mid] < nums[mid+1]:
l = mid + 1
else:
r = mid
return l
# 59/59 cases passed (44 ms)
# Your runtime beats 72.82 % of python3 submissions
# Your memory usage beats 100 % of python3 submissions (12.8 MB)


## 找局部最小值

class Solution {
public:
int getLessIndex(vector<int> arr) {
int n = arr.size();
if(n == 0)
{
return -1;
}
if(n == 1 || arr[0] < arr[1])
{
return 0;
}
if(arr[n-1] < arr[n-2])
{
return n-1;
}
int l = 1;
int r = n - 2;
int mid = -1;
while(l <= r)
{
mid = l + (r - l) / 2;
if(arr[mid-1] > arr[mid] && arr[mid] < arr[mid+1])
{
return mid;
}
else if(arr[mid-1] < arr[mid])
{
r = mid - 1;
}
else
{
l = mid + 1;
}
}
return -1;
}
};