162. 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 #

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
}