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# Find Peak Element

```
There is an integer array which has the following features:

The numbers in adjacent positions are different.
A[0] < A[1] && A[A.length - 2] > A[A.length - 1].
We define a position P is a peak if:

A[P] > A[P-1] && A[P] > A[P+1]
Find a peak element in this array. Return the index of the peak.

It's guaranteed the array has at least one peak.
The array may contain multiple peeks, find any of them.
The array has at least 3 numbers in it.
Have you met this question in a real interview?  
Example
Given [1, 2, 1, 3, 4, 5, 7, 6]

Return index 1 (which is number 2) or 6 (which is number 7)

Challenge
Time complexity O(logN)
```

```java
public class Solution {
    /*
     * @param A: An integers array.
     * @return: return any of peek positions.
     */
    public int findPeak(int[] A) {
        int start = 1, end = A.length -2;

        while(start +1 < end){
            int mid = start+(end - start)/2;

            if (A[mid]> A[mid -1] && A[mid] > A[mid]+1) {
                end = mid;
            }

            else if(A[mid] < A[mid-1]){
                end = mid;
            }else{
                start = mid;
            }
        }


        if (A[start] > A[end]) {
            return start;
        }else{
            return end;
        }
    }
}
```

二刷code

```java
class Solution {
    public int findPeakElement(int[] nums) {
        
        
        
        if(nums == null || nums.length == 0){
            return -1;
        }
        
        int start = 0, end = nums.length - 1;
        
        while(start + 1 < end){
            int mid = start + (end - start)/2;
            
            if(nums[mid] > nums[mid+1]){
                end = mid;
            }else{
                start = mid;
            }
        }
        
        if(nums[start] > nums[end]){
            return start;
        }else{
            return end;
        }
    }
}
```