[LeetCode] (medium) 162. Find Peak Element

最新推荐文章于 2020-11-25 09:21:46 发布

Frontier_Setter

最新推荐文章于 2020-11-25 09:21:46 发布

阅读量129

点赞数

CC 4.0 BY-SA版权

分类专栏： LeetCode 二分文章标签： LeetCode

本文链接：https://blog.youkuaiyun.com/Frontier_Setter/article/details/87468572

LeetCode 同时被 2 个专栏收录

98 篇文章

订阅专栏

二分

11 篇文章

订阅专栏

本文探讨了在给定数组中寻找峰值元素的问题，峰值元素定义为大于其邻居的元素。文章提供了两种解决方案，一种是线性方法，通过遍历数组来找到峰值；另一种是对数方法，采用二分查找策略，在对数时间内定位峰值位置。通过对两种方法的对比，读者可以理解不同复杂度下解决问题的策略。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

https://leetcode.com/problems/find-peak-element/

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.

线性方法：

很直观，直接遍历寻找不满足单调性的位置

class Solution {
public:
    int findPeakElement(vector<int>& nums) {
        for(int i = 1; i < nums.size(); ++i){
            if(nums[i] > nums[i-1]) continue;
            else return i-1;
        }
        return nums.size()-1;
    }
};

对数方法：

要求是对数时间，又是查找，考虑的就是二分查找。类比普通的二分查找，这道题的二分不变性就应该是“peak点始终保持在搜索区间内”

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