# [162. Find Peak Element](https://leetcode.com/problems/find-peak-element/)


## 题目

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.

## 题目大意

峰值元素是指其值大于左右相邻值的元素。给定一个输入数组 nums，其中 nums[i] ≠ nums[i+1]，找到峰值元素并返回其索引。数组可能包含多个峰值，在这种情况下，返回任何一个峰值所在位置即可。你可以假设 nums[-1] = nums[n] = -∞。

说明:

- 你的解法应该是 O(logN) 时间复杂度的。


## 解题思路

- 给出一个数组，数组里面存在多个“山峰”，(山峰的定义是，下标 `i` 比 `i-1`、`i+1` 位置上的元素都要大)，找到这个“山峰”，并输出其中一个山峰的下标。
- 这一题是第 852 题的伪加强版，第 852 题中只存在一个山峰，这一题存在多个山峰。但是实际上搜索的代码是一样的，因为此题只要求随便输出一个山峰的下标即可。思路同第 852 题。


## 代码

```go

package leetcode

// 解法一 二分
func findPeakElement(nums []int) int {
	if len(nums) == 0 || len(nums) == 1 {
		return 0
	}
	low, high := 0, len(nums)-1
	for low <= high {
		mid := low + (high-low)>>1
		if (mid == len(nums)-1 && nums[mid-1] < nums[mid]) || (mid > 0 && nums[mid-1] < nums[mid] && (mid <= len(nums)-2 && nums[mid+1] < nums[mid])) || (mid == 0 && nums[1] < nums[0]) {
			return mid
		}
		if mid > 0 && nums[mid-1] < nums[mid] {
			low = mid + 1
		}
		if mid > 0 && nums[mid-1] > nums[mid] {
			high = mid - 1
		}
		if mid == low {
			low++
		}
		if mid == high {
			high--
		}
	}
	return -1
}

// 解法二 二分
func findPeakElement1(nums []int) int {
	low, high := 0, len(nums)-1
	for low < high {
		mid := low + (high-low)>>1
		// 如果 mid 较大，则左侧存在峰值，high = m，如果 mid + 1 较大，则右侧存在峰值，low = mid + 1
		if nums[mid] > nums[mid+1] {
			high = mid
		} else {
			low = mid + 1
		}
	}
	return low
}

```