给定链表中峰之间的最大距离
给定一个长度为N的链表lis ，任务是确定给定链表的两个连续峰值之间的最大距离。峰值被定义为具有大于其邻居的值的节点。两个节点之间的距离定义为它们之间存在的节点数。
例子：
Input: lis = 1 -> 2 -> 3 -> 1 -> 5 -> 4 -> 4 -> 10 -> 7
Output: 2
Explanation: The peaks in the linkedlist are 3, 5, 10
The distance between 3 and 5 is 1.
The distance between 5 and 10 is 2.
The maximum distance is 2.
Input: lis = 1 -> 3 -> 1 -> 1 ->1 -> 1 -> 4 -> 2 -> 7
Output: 4
Explanation: The peaks in the linkedlist are 3, 4, 7
The distance between 3 and 4 is 4.
The distance between 4 and 7 is 1.
The maximum distance is 4.
方法：解决方案基于贪婪方法。请按照以下步骤解决问题：
- 创建一个空集来存储链表中的峰值节点。
- 遍历链表并找到峰值节点。
- 再次遍历链表并检查当前节点是否为峰值。如果是，则计算最后一个峰值节点和当前节点之间的距离。
下面是上述方法的实现。
C++
// C++ code to implement the above approach
#include
using namespace std;
// Class(struct) for the structure of a node
struct Node {
int data;
Node* next;
Node(int data, Node* next)
: data(data)
, next(next)
{
}
};
// Function to find the maximum distance
void findMaxGap(Node* head)
{
// Create a set to store
// all the peak nodes
set peaks;
// Pre points to previous of current node
Node* pre = new Node(-1, head);
// Current is initially head
Node* cur = head;
// Loop till current is not NULL
while (cur != NULL) {
// Find next of current
Node* next = cur->next;
// One node only
if (pre->next == head && next == NULL)
peaks.insert(cur);
// For the 1st node check only next
else if (pre->next == head
&& next->data < cur->data)
peaks.insert(cur);
// For the last node check
// only the prev node
else if (next == NULL && pre->data < cur->data)
peaks.insert(cur);
// Check prev and next nodes for all
// intermediate nodes
else if (pre->data < cur->data
&& next->data < cur->data)
peaks.insert(cur);
// Move pre and cur pointer
pre = pre->next;
cur = cur->next;
}
// Initialize
int lastPeakIndex = -1, currentIndex = 0, maxGap = 0,
gap = 0;
cur = head;
// Iterate through the linkedlist
while (cur != NULL) {
// If current node is a peak
if (peaks.count(cur)) {
// If current node is first peak
// then update lastPeakIndex
// and move ahead
if (lastPeakIndex == -1)
lastPeakIndex = currentIndex;
// If current node peak then
// calculate gap and update
// lastPeakIndex and move ahead
else {
gap = currentIndex - lastPeakIndex - 1;
maxGap = max(gap, maxGap);
lastPeakIndex = currentIndex;
}
}
// Increment currentIndex
// move current node ahead
currentIndex++;
cur = cur->next;
}
cout << maxGap << "\n";
}
// Driver code
int main()
{
// Create the linkedlist
Node* head = new Node(1, NULL);
head->next = new Node(2, NULL);
head->next->next = new Node(3, NULL);
head->next->next->next = new Node(1, NULL);
head->next->next->next->next = new Node(5, NULL);
head->next->next->next->next->next = new Node(4, NULL);
head->next->next->next->next->next->next
= new Node(4, NULL);
head->next->next->next->next->next->next->next
= new Node(10, NULL);
head->next->next->next->next->next->next->next->next
= new Node(7, NULL);
// Function call
findMaxGap(head);
}
// This code is contributed by Taranpreet Java
// Java code to implement the above approach
import java.io.*;
import java.util.*;
// Class for the structure of a node
class Node {
int data;
Node next;
public Node(int data, Node next)
{
this.data = data;
this.next = next;
}
}
class GFG {
// Driver code
public static void main(String[] args)
{
// Create the linkedlist
Node head = new Node(1, null);
head.next = new Node(2, null);
head.next.next = new Node(3, null);
head.next.next.next = new Node(1, null);
head.next.next.next.next = new Node(5, null);
head.next.next.next.next.next = new Node(4, null);
head.next.next.next.next.next.next
= new Node(4, null);
head.next.next.next.next.next.next.next
= new Node(10, null);
head.next.next.next.next.next.next.next.next
= new Node(7, null);
// Function call
findMaxGap(head);
}
// Function to find the maximum distance
static void findMaxGap(Node head)
{
// Create a set to store
// all the peak nodes
Set peaks = new HashSet<>();
// Pre points to previous of current node
Node pre = new Node(-1, head);
// Current is initially head
Node cur = head;
// Loop till current is not null
while (cur != null) {
// Find next of current
Node next = cur.next;
// One node only
if (pre.next == head && next == null)
peaks.add(cur);
// For the 1st node check only next
else if (pre.next == head
&& next.data < cur.data)
peaks.add(cur);
// For the last node check
// only the prev node
else if (next == null
&& pre.data < cur.data)
peaks.add(cur);
// Check prev and next nodes for all
// intermediate nodes
else if (pre.data < cur.data
&& next.data < cur.data)
peaks.add(cur);
// Move pre and cur pointer
pre = pre.next;
cur = cur.next;
}
// Initialize
int lastPeakIndex = -1,
currentIndex = 0,
maxGap = 0, gap = 0;
cur = head;
// Iterate through the linkedlist
while (cur != null) {
// If current node is a peak
if (peaks.contains(cur)) {
// If current node is first peak
// then update lastPeakIndex
// and move ahead
if (lastPeakIndex == -1)
lastPeakIndex = currentIndex;
// If current node peak then
// calculate gap and update
// lastPeakIndex and move ahead
else {
gap = currentIndex
- lastPeakIndex - 1;
maxGap = Math.max(gap, maxGap);
lastPeakIndex = currentIndex;
}
}
// Increment currentIndex
// move current node ahead
currentIndex++;
cur = cur.next;
}
System.out.println(maxGap);
}
} C#
// C# code to implement the above approach
using System;
using System.Collections.Generic;
// Class for the structure of a node
class Node {
public int data;
public Node next;
public Node(int data, Node next)
{
this.data = data;
this.next = next;
}
}
public class GFG {
// Driver code
public static void Main(String[] args)
{
// Create the linkedlist
Node head = new Node(1, null);
head.next = new Node(2, null);
head.next.next = new Node(3, null);
head.next.next.next = new Node(1, null);
head.next.next.next.next = new Node(5, null);
head.next.next.next.next.next = new Node(4, null);
head.next.next.next.next.next.next
= new Node(4, null);
head.next.next.next.next.next.next.next
= new Node(10, null);
head.next.next.next.next.next.next.next.next
= new Node(7, null);
// Function call
findMaxGap(head);
}
// Function to find the maximum distance
static void findMaxGap(Node head)
{
// Create a set to store
// all the peak nodes
HashSet peaks = new HashSet();
// Pre points to previous of current node
Node pre = new Node(-1, head);
// Current is initially head
Node cur = head;
// Loop till current is not null
while (cur != null) {
// Find next of current
Node next = cur.next;
// One node only
if (pre.next == head && next == null)
peaks.Add(cur);
// For the 1st node check only next
else if (pre.next == head
&& next.data < cur.data)
peaks.Add(cur);
// For the last node check
// only the prev node
else if (next == null
&& pre.data < cur.data)
peaks.Add(cur);
// Check prev and next nodes for all
// intermediate nodes
else if (pre.data < cur.data
&& next.data < cur.data)
peaks.Add(cur);
// Move pre and cur pointer
pre = pre.next;
cur = cur.next;
}
// Initialize
int lastPeakIndex = -1,
currentIndex = 0,
maxGap = 0, gap = 0;
cur = head;
// Iterate through the linkedlist
while (cur != null) {
// If current node is a peak
if (peaks.Contains(cur)) {
// If current node is first peak
// then update lastPeakIndex
// and move ahead
if (lastPeakIndex == -1)
lastPeakIndex = currentIndex;
// If current node peak then
// calculate gap and update
// lastPeakIndex and move ahead
else {
gap = currentIndex
- lastPeakIndex - 1;
maxGap = Math.Max(gap, maxGap);
lastPeakIndex = currentIndex;
}
}
// Increment currentIndex
// move current node ahead
currentIndex++;
cur = cur.next;
}
Console.WriteLine(maxGap);
}
}
// This code is contributed by 29AjayKumar 输出
2时间复杂度： O(N)
辅助空间： O(N)