Find Peak Element

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最新推荐文章于 2024-11-13 02:34:54 发布

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分类专栏： leetcode leetCode 文章标签： leetcode binary search

本文链接：https://blog.youkuaiyun.com/qingtian_1993/article/details/81010489

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本文介绍了一种使用二分查找法寻找峰值元素的方法。峰值元素是指比其邻居大的元素，在给定的数组中可能存在多个峰值，只需返回任一峰值的索引。文章提供了具体的实现代码及时间复杂度分析。

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Find Peak Element

Description

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.

Tags: Binary Search

解读题意

给定一个数组，可能会存在多个顶峰，即nums[i] > nums[i+1]，找出其中一个顶峰即可，并返回顶峰的下标

思路1

利用二叉搜索树来解决问题。该题的数组nums[]本身是无序的，但是我们不用返回数组的数值，可以将数组看作nums[true,false,true,...,true,false]，判断nums[i] < nums[i+1]即可。

public class Solution1 {

    public int findPeakElement(int[] nums) {

        if (nums == null || nums.length == 0)
            throw new IllegalArgumentException();

        int left = 0, right = nums.length - 1;
        while (left < right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] < nums[mid + 1])
                left = mid + 1;
            else
                right = mid;
        }

        return left;
    }

}