LeetCode-Find Peak Element

最新推荐文章于 2024-11-13 02:34:54 发布

转载最新推荐文章于 2024-11-13 02:34:54 发布 · 66 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/incrediblechangshuo/p/4319589.html

本文介绍了一种在给定数组中寻找峰值元素的方法，并给出了一个具体的Java实现。峰值元素是指大于其邻居的元素。文章通过二分查找的方式实现了在对数时间内找到峰值元素的目标。

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.

click to show spoilers.

Note:
Your solution should be in logarithmic complexity.

这题的关键是弄清楚为什么可以使用二分法。对于两端的节点，因为num[-1] = num[n] = -∞，所以只要数组中没有连续出现的相同元素，这个peak element就总是存在的。

二分查找所依据的关键，就是如果num[mid]<=num[mid-1]，那么peak element一定存在于区间[low, mid-1]；如果num[mid]<=num[mid+1]，那么peak element一定存在于区间[mid+1, high].

public class Solution {
    public int findPeakElement(int[] num) {
        int len = num.length;
        if( len <= 1){
            return 0;
        }
        if(num[0] > num[1]){
            return 0;
        }
        if(num[len-1] > num[len-2]){
            return len-1;
        }
        
        int low = 0;
        int high = len-1;
        
        while(high>low){
            int mid = num[(low+high)/2];
            if( mid>num[(low+high)/2+1] && mid>num[(low+high)/2-1] ){
                return (low+high)/2;
            }
            else if(mid < num[(low+high)/2+1]){
                low = (low+high)/2;
                
            }
            
            else if(mid <num[(low+high)/2-1]){
                high = (low+high)/2;
            }
            
        }
        return -1;
        
        
    }
    
  
    
}

转载于:https://www.cnblogs.com/incrediblechangshuo/p/4319589.html