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A peak element is an element that is strictly greater than its neighbors.

Given a **0-indexed** integer array `nums`

, find a peak element, and return its index. If the array contains multiple peaks, return the index to **any of the peaks**.

You may imagine that `nums[-1] = nums[n] = -∞`

. In other words, an element is always considered to be strictly greater than a neighbor that is outside the array.

You must write an algorithm that runs in `O(log n)`

time.

**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: 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.
```

**Constraints:**

`1 <= nums.length <= 1000`

`-2`

^{31}<= nums[i] <= 2^{31}- 1`nums[i] != nums[i + 1]`

for all valid`i`

.

```
public int findPeakElement(int[] nums) {
int N = nums.length;
if (N == 1) {
return 0;
}
int left = 0, right = N - 1;
while (right - left > 1) {
int mid = left + (right - left) / 2;
if (nums[mid] < nums[mid + 1]) {
left = mid + 1;
} else {
right = mid;
}
}
return (left == N - 1 || nums[left] > nums[left + 1]) ? left : right;
}
```

```
def findPeakElement(self, nums):
left = 0
right = len(nums)-1
# handle condition 3
while left < right-1:
mid = (left+right)/2
if nums[mid] > nums[mid+1] and nums[mid] > nums[mid-1]:
return mid
if nums[mid] < nums[mid+1]:
left = mid+1
else:
right = mid-1
#handle condition 1 and 2
return left if nums[left] >= nums[right] else right
```

```
class Solution {
public:
int findPeakElement(vector<int>& nums) {
int lo = 0, hi = nums.size()-1, mid;
while (lo < hi) {
mid = lo + (hi - lo) / 2;
if (nums[mid] < nums[mid + 1])
lo = mid + 1;
else hi = mid;
}
return lo;
}
};
```

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