【LeetCode】162. Find Peak Element-优快云博客

本文链接：https://blog.youkuaiyun.com/u012352249/article/details/49202287

本文介绍了一种寻找峰值元素的方法，峰值元素是指大于其相邻元素的元素。对于输入数组num[i]不等于num[i+1]的情况，该算法能有效地找到峰值元素并返回其索引。文章提供了详细的Java实现代码。

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最近去某公司面试，笔试题中有这个题目，整理，记录

问题描述

A peak element is an element that is greater than its neighbors.

Given an input array where num[i] ≠ num[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 num[-1] = num[n] = -∞.

For example, in array [1, 2, 3, 1], 3 is a peak element and your function should return the index number 2.

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My Solution

public class Solution {
    public int findPeakElement(int[] nums) {
        if(nums.length==1)  //只有一个元素
            return 0;
        else if(nums.length==2)//有两个元素，返回大的元素的索引
        {
            if(nums[0]>nums[1])
                return 0;
            else if(nums[0]<nums[1])
            {
                return 1;
            }else
                return -1;

        }else{//元素个数大于2
            for(int i=0;i<nums.length;i++){
                if(i==nums.length-1){   //遍历到最后一个元素
                    if(nums[i]>nums[i-1] ){
                        return i;
                    }
                }else if(i==0){      //第一个元素
                    if(nums[i]>nums[i+1]){
                          return i;
                      }

                }else{    //中间元素的情况
                        //i的左右邻居都与i相等
                        if(nums[i]==nums[i-1]||nums[i]==nums[i+1])
                            continue;
                        //i-1<i<i+1继续寻找
                        else if(nums[i]>nums[i-1]&&nums[i]<nums[i+1])
                            continue;
                        else
                            return i;
                }
        }  
    }
    return -1;
    }
}