【LeetCode & 剑指offer刷题】查找与排序题9：Find Peak Element-优快云博客

本文探讨了在数组中寻找峰值元素的两种方法，一种是通过遍历找到最大值，另一种是采用二分查找思想提高效率。峰值元素是指大于其邻居的元素，文章提供了C++代码实现，并解释了算法复杂度。

【LeetCode & 剑指offer 刷题笔记】目录（持续更新中...）

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.

C++

    //问题：找极大值元素（返回任意一个极大值均可）
   

    /*
   

    方法一：找最大值一定是极大值
   

    O(n)
   

    */
   

    class 
    Solution
   

    {
   

    public
    :
   

        
    int 
    findPeakElement
    (
    vector
    <
    int
    >& 
    nums
    )
   

        
    {
   

            
    int 
    n 
    = 
    nums
    .
    size
    ();
   

            
    if
    (
    n
    ==
    0
    ) 
    return 
    -
    1
    ; 
    //为空时，返回-1
   

            
    int 
    max_index 
    = 
    0
    ;
   

            
    for
    (
    int 
    i 
    = 
    0
    ; 
    i 
    < 
    n
    ; 
    i
    ++)
    //找序列中的最大值，一定是极大值
   

            
    {
   

                
    if
    (
    nums
    [
    i
    ]>
    nums
    [
    max_index
    ]) 
    max_index 
    = 
    i
    ;
   

            
    }
   

            
    return 
    max_index
    ;
   

        
    }
   

    };
   

     
   

    /*
    掌握
   

    方法二：借助二分查找思路
   

    right指的数比后一个数大，left指的数比前一个数大，当两个“指针”相遇时，就可以满足极大值条件（两个指针按二分跨度走，故效率较高）
   

    有点像lower_bound函数
   

    O(logn)
   

    */
   

    class 
    Solution
   

    {
   

    public
    :
   

        
    int 
    findPeakElement
    (
    vector
    <
    int
    >& 
    nums
    )
   

        
    {
   

            
    if
    (
    nums
    .
    empty
    () 
    return 
    -
    1
    ; 
    //
    为空时，返回
    -1
   

           
   

            
    int 
    left 
    = 
    0
    ,
    right 
    = 
    nums
    .
    size
    ()-
    1
    ;
   

            
    while
    (
    left
    <
    right
    ) 
    //left = right
    时退出循环，结果是
    left,right,mid
    均指向极大值位置
   

            
    {
   

                
    int 
    mid 
    = 
    (
    left
    +
    right
    )/
    2
    ;
   

                
    if
    (
    nums
    [
    mid
    ] 
    < 
    nums
    [
    mid
    +
    1
    ]) 
    //如果中间值比右边值小，说明峰值在右边
   

                    left 
    = 
    mid
    +
    1
    ;
   

                
    else 
    //
    如果中间值比右边值大，说明峰值在左边（包括
    a[mid]
    ，故取
    right=mid
    ）
   

                    right 
    = 
    mid
    ;
   

            
    }
   

          
   

            
    return right;
   

          
   

        
    }
   

    };
   