# Find Peak Element

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.

## Solution

Linear solution that check whether the next element is smaller, when it is, peak was found.

```
public class Solution {
    public int findPeakElement(int[] nums) {
        for (int i = 0; i < nums.length - 1; i++) {
            if (nums[i] > nums[i + 1])
                return i;
        }
        return nums.length - 1;
    }
}
```

Iterative binary search.

```
class Solution {
    public int findPeakElement(int[] nums) {
        int start = 0;
        int end = nums.length - 1;
        while (start < end) {
            int middle = start + ((end - start) / 2);
            int middleVal = nums[middle];
            // 1, 2, 3, 1
            if (middleVal > nums[middle + 1]) {
                end = middle;
            } else {
                start = middle + 1;
            }
        }
        return start;
    }
}
```

Recursive binary search.

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


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