# Find Peak Element

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

> Given an input array where num[i] ≠ num[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 num[-1] = num[n] = -∞.

> For example, in array [1, 2, 3, 1], 3 is a peak element and your function should return the index number 2.

这题要求我们在一个无序的数组里面找到一个peak元素,所谓peak,就是值比两边邻居大就行了。

对于这题,最简单地解法就是遍历数组,只要找到第一个元素,大于两边就可以了,复杂度为O(N)。但这题还可以通过二分来做。

首先我们找到中间节点mid,如果大于两边返回当前index就可以了,如果左边的节点比mid大,那么我们可以继续在左半区间查找,这里面一定存在一个peak,为什么这么说呢?假设此时的区间范围为[0, mid - 1], 因为num[mid - 1]一定大于num[mid]了,如果num[mid - 2] <= num[mid - 1],那么num[mid - 1]就是一个peak。如果num[mid - 2] > num[mid - 1],那么我们就继续在[0, mid - 2]区间查找,因为num[-1]为负无穷,所以最终我们绝对能在左半区间找到一个peak。同理右半区间一样。

代码如下:

```c++
class Solution {
public:
int findPeakElement(const vector &num) {
int n = num.size();
if(n == 1) {
return 0;
}

int start = 0;
int end = n - 1;
int mid = 0;

while(start <= end) {
mid = start + (end - start) / 2;
if((mid == 0 || num[mid] >= num[mid - 1]) &&
(mid == n - 1 || num[mid] >= num[mid + 1])) {
return mid;
}else if(mid > 0 && num[mid-1] > num[mid]) {
end = mid - 1;
} else {
start = mid + 1;
}
}
return mid;
}
};
```