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94 lines
2.6 KiB
Markdown
Executable File
94 lines
2.6 KiB
Markdown
Executable File
# [162. Find Peak Element](https://leetcode.com/problems/find-peak-element/)
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## 题目
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A peak element is an element that is greater than its neighbors.
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Given an input array `nums`, where `nums[i] ≠ nums[i+1]`, find a peak element and return its index.
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The array may contain multiple peaks, in that case return the index to any one of the peaks is fine.
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You may imagine that `nums[-1] = nums[n] = -∞`.
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**Example 1**:
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Input: nums = [1,2,3,1]
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Output: 2
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Explanation: 3 is a peak element and your function should return the index number 2.
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**Example 2**:
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Input: nums = [1,2,1,3,5,6,4]
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Output: 1 or 5
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Explanation: Your function can return either index number 1 where the peak element is 2,
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or index number 5 where the peak element is 6.
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**Note**:
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Your solution should be in logarithmic complexity.
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## 题目大意
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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] = -∞。
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说明:
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- 你的解法应该是 O(logN) 时间复杂度的。
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## 解题思路
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- 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。
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- 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。
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## 代码
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```go
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package leetcode
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// 解法一 二分
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func findPeakElement(nums []int) int {
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if len(nums) == 0 || len(nums) == 1 {
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return 0
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}
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low, high := 0, len(nums)-1
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for low <= high {
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mid := low + (high-low)>>1
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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]) {
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return mid
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}
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if mid > 0 && nums[mid-1] < nums[mid] {
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low = mid + 1
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}
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if mid > 0 && nums[mid-1] > nums[mid] {
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high = mid - 1
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}
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if mid == low {
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low++
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}
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if mid == high {
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high--
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}
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}
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return -1
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}
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// 解法二 二分
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func findPeakElement1(nums []int) int {
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low, high := 0, len(nums)-1
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for low < high {
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mid := low + (high-low)>>1
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// 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
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if nums[mid] > nums[mid+1] {
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high = mid
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} else {
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low = mid + 1
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}
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}
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return low
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}
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``` |