162. Find Peak Element

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

 这个题就是比较 mid 的时候， 和 mid + 1 index 的value 比较， 如果 mid + 1 的 value 比较大的话， 那么 local peak 肯定在 右边. it also covers the corner case , like 1 2 , then it returns index 1 

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.

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