在排序的链表中查找中位数
原文：https://www.geeksforgeeks.org/finding-median-in-a-sorted-linked-list/
给定N元素的排序链表。 任务是在给定的排序链表中找到中位数。
我们知道排序数组中的中位数是中间元素。
查找N个排序数字中位数的过程：
if N is odd:
median is N/2th element
else
median is N/2th element + (N/2+1)th element
示例：
Input : 1->2->3->4->5->NULL
Output : 3
Input : 1->2->3->4->5->6->NULL
Output : 3.5
简单方法
-
遍历链表并计算所有元素。
-
如果
count是奇数，则再次遍历链表并找到第n / 2个元素。 -
如果
count是偶数，则再次遍历链表并找到：（第
n / 2个元素+第n / 2 + 1个元素）/ 2
注意：上述解决方案两次遍历链表。
有效方法：一种有效方法是使用两个指针遍历列表以查找元素数。 参见此帖子的方法 2。
我们可以使用上述算法来找到链表的中位数。 使用此算法，我们无需计算元素的数量：
-
如果
fast_ptr不为NULL，则意味着链表包含奇数元素，我们仅打印slow_ptr的数据。 -
否则，如果
fast_ptr达到NULL，则意味着链表包含偶数元素，我们创建了slow_ptr的前一个节点的备份并打印(pre_of_slow->data + slow_ptr->data) / 2
下面是上述方法的实现：
C++
// C++ program to find median
// of a linked list
#include <bits/stdc++.h>
using namespace std;
// Link list node
struct Node {
int data;
struct Node* next;
};
/* Function to get the median of the linked list */
void printMidean(Node* head)
{
Node* slow_ptr = head;
Node* fast_ptr = head;
Node* pre_of_slow = head;
if (head != NULL) {
while (fast_ptr != NULL && fast_ptr->next != NULL) {
fast_ptr = fast_ptr->next->next;
// previous of slow_ptr
pre_of_slow = slow_ptr;
slow_ptr = slow_ptr->next;
}
// if the below condition is true linked list
// contain odd Node
// simply return middle element
if (fast_ptr != NULL)
cout << "Median is : " << slow_ptr->data;
// else linked list contain even element
else
cout << "Median is : "
<< float(slow_ptr->data + pre_of_slow->data) / 2;
}
}
/* Given a reference (pointer to
pointer) to the head of a list
and an int, push a new node on
the front of the list. */
void push(struct Node** head_ref, int new_data)
{
// allocate node
Node* new_node = new Node;
// put in the data
new_node->data = new_data;
// link the old list
// off the new node
new_node->next = (*head_ref);
// move the head to point
// to the new node
(*head_ref) = new_node;
}
// Driver Code
int main()
{
// Start with the
// empty list
struct Node* head = NULL;
// Use push() to construct
// below list
// 1->2->3->4->5->6
push(&head, 6);
push(&head, 5);
push(&head, 4);
push(&head, 3);
push(&head, 2);
push(&head, 1);
// Check the count
// function
printMidean(head);
return 0;
}
Java
// Java program to find median
// of a linked list
class GFG
{
// Link list node
static class Node
{
int data;
Node next;
};
/* Function to get the median of the linked list */
static void printMidean(Node head)
{
Node slow_ptr = head;
Node fast_ptr = head;
Node pre_of_slow = head;
if (head != null)
{
while (fast_ptr != null && fast_ptr.next != null)
{
fast_ptr = fast_ptr.next.next;
// previous of slow_ptr
pre_of_slow = slow_ptr;
slow_ptr = slow_ptr.next;
}
// if the below condition is true linked list
// contain odd Node
// simply return middle element
if (fast_ptr != null)
{
System.out.print("Median is : " + slow_ptr.data);
}
// else linked list contain even element
else
{
System.out.print("Median is : "
+ (float) (slow_ptr.data + pre_of_slow.data) / 2);
}
}
}
/* Given a reference (pointer to
pointer) to the head of a list
and an int, push a new node on
the front of the list. */
static Node push(Node head_ref, int new_data)
{
// allocate node
Node new_node = new Node();
// put in the data
new_node.data = new_data;
// link the old list
// off the new node
new_node.next = head_ref;
// move the head to point
// to the new node
head_ref = new_node;
return head_ref;
}
// Driver Code
public static void main(String[] args)
{
// Start with the
// empty list
Node head = null;
// Use push() to construct
// below list
// 1.2.3.4.5.6
head = push(head, 6);
head = push(head, 5);
head = push(head, 4);
head = push(head, 3);
head = push(head, 2);
head = push(head, 1);
// Check the count
// function
printMidean(head);
}
}
// This code is contributed by PrinciRaj1992
C
// C# program to find median
// of a linked list
using System;
class GFG
{
// Link list node
class Node
{
public int data;
public Node next;
};
/* Function to get the median
of the linked list */
static void printMidean(Node head)
{
Node slow_ptr = head;
Node fast_ptr = head;
Node pre_of_slow = head;
if (head != null)
{
while (fast_ptr != null &&
fast_ptr.next != null)
{
fast_ptr = fast_ptr.next.next;
// previous of slow_ptr
pre_of_slow = slow_ptr;
slow_ptr = slow_ptr.next;
}
// if the below condition is true linked list
// contain odd Node
// simply return middle element
if (fast_ptr != null)
{
Console.Write("Median is : " +
slow_ptr.data);
}
// else linked list contain even element
else
{
Console.Write("Median is : " +
(float)(slow_ptr.data +
pre_of_slow.data) / 2);
}
}
}
/* Given a reference (pointer to
pointer) to the head of a list
and an int, push a new node on
the front of the list. */
static Node push(Node head_ref, int new_data)
{
// allocate node
Node new_node = new Node();
// put in the data
new_node.data = new_data;
// link the old list
// off the new node
new_node.next = head_ref;
// move the head to point
// to the new node
head_ref = new_node;
return head_ref;
}
// Driver Code
public static void Main(String[] args)
{
// Start with the
// empty list
Node head = null;
// Use push() to construct
// below list
// 1->2->3->4->5->6
head = push(head, 6);
head = push(head, 5);
head = push(head, 4);
head = push(head, 3);
head = push(head, 2);
head = push(head, 1);
// Check the count
// function
printMidean(head);
}
}
// This code is contributed by Rajput-Ji
输出：
Median is : 3.5
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