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.

用binary search。如果nums[i] > nums[i+1]，說明peak在左半部分，r = mid；如果nums[i] <= nums[i+1]，說明peak在右半部分，l = mid + 1

時間：O(log2(N))，空間：O(1)

class Solution {
    public int findPeakElement(int[] nums) {
        int l = 0, r = nums.length - 1;
        
 
while(l < r) {
            int mid = l + (r - l) / 2;
            if(nums[mid] > nums[mid + 1])
                r = mid;
            else
                l = mid + 1;
        }
        return l;
    }
}