在链表中查找频率至少为 N/3 的元素

给定一个由字符串作为节点值的大小为N的链表，任务是在链表中找到频率大于[N/3]的多数字符串。

注意：保证只有一个多数字符串。

例子：

Input: head -> geeks -> geeks -> abcd -> game -> knight -> geeks -> harry. 
Output: geeks. 
Explanation: 
The frequency of geeks string in link list is 3 which is greater than [7/3] i.e 2.

Input: head -> hot -> hot -> cold -> hot -> hot 
Output: hot 
Explanation: 
The frequency of hot string in the link list is 4 which is greater than [5/3] i.e 1. 

天真的方法：
将每个字符串的频率存储在 Map 中。遍历地图并查找频率≥ N / 3的字符串。

时间复杂度： O(N)
辅助空间： O(N)

有效的方法：
这个想法是基于摩尔的投票算法。
找到两个候选人并检查这两个候选人中的任何一个是否实际上是多数元素。

下面是上述方法的实现：

C++

// C++ program to find an
// element with frequency
// of at least N / 3
// in a linked list
 
#include 
using namespace std;
 
// Structure of a node
// for the linked list
struct node {
    string i;
    node* next = NULL;
};
 
// Utility function to
// create a node
struct node* newnode(string s)
{
    struct node* temp = (struct node*)
        malloc(sizeof(struct node));
    temp->i = s;
    temp->next = NULL;
    return temp;
}
 
// Function to find and return
// the element with frequency
// of at least N/3
string Majority_in_linklist(node* head)
{
    // Candidates for
    // being the required
    // majority element
    string s = "", t = "";
 
    // Store the frequencies
    // of the respective candidates
    int p = 0, q = 0;
    node* ptr = NULL;
 
    // Iterate all nodes
    while (head != NULL) {
        if (s.compare(head->i) == 0) {
            // Increase frequency
            // of candidate s
            p = p + 1;
        }
        else {
            if (t.compare(head->i) == 0) {
                // Increase frequency
                // of candidate t
                q = q + 1;
            }
            else {
                if (p == 0) {
                    // Set the new sting as
                    // candidate for majority
                    s = head->i;
                    p = 1;
                }
                else {
                    if (q == 0) {
                        // Set the new sting as
                        // second candidate
                        // for majority
                        t = head->i;
                        q = 1;
                    }
                    else {
                        // Decrease the frequency
                        p = p - 1;
                        q = q - 1;
                    }
                }
            }
        }
        head = head->next;
    }
    head = ptr;
    p = 0;
    q = 0;
 
    // Check the frequency of two
    // final selected candidate linklist
    while (head != NULL) {
        if (s.compare(head->i) == 0) {
            // Increase the frequency
            // of first candidate
            p = 1;
        }
        else {
            if (t.compare(head->i) == 0) {
                // Increase the frequency
                // of second candidate
                q = 1;
            }
        }
        head = head->next;
    }
    // Return the string with
    // higher frequency
    if (p > q) {
        return s;
    }
    else {
        return t;
    }
}
// Driver Code
int main()
{
    node* ptr = NULL;
    node* head = newnode("geeks");
    head->next = newnode("geeks");
    head->next->next = newnode("abcd");
    head->next->next->next
        = newnode("game");
    head->next->next->next->next
        = newnode("game");
    head->next->next->next->next->next
        = newnode("knight");
    head->next->next->next->next->next->next
        = newnode("harry");
    head->next->next->next->next->next->next
        ->next
        = newnode("geeks");
 
    cout << Majority_in_linklist(head) << endl;
 
    return 0;
}

Java

// Java program to find an
// element with frequency
// of at least N / 3
// in a linked list
class GFG{
 
// Structure of a node
// for the linked list
static class node
{
    String i;
    node next = null;
};
 
// Utility function to
// create a node
static node newnode(String s)
{
    node temp = new node();
    temp.i = s;
    temp.next = null;
    return temp;
}
 
// Function to find and return
// the element with frequency
// of at least N/3
static String Majority_in_linklist(node head)
{
     
    // Candidates for
    // being the required
    // majority element
    String s = "";
    String t = "";
     
    // Store the frequencies
    // of the respective candidates
    int p = 0, q = 0;
    node ptr = null;
 
    // Iterate all nodes
    while (head != null)
    {
        if (s.equals(head.i))
        {
             
            // Increase frequency
            // of candidate s
            p = p + 1;
        }
        else
        {
            if (t.equals(head.i))
            {
                 
                // Increase frequency
                // of candidate t
                q = q + 1;
            }
            else
            {
                if (p == 0)
                {
                     
                    // Set the new sting as
                    // candidate for majority
                    s = head.i;
                    p = 1;
                }
                else
                {
                    if (q == 0)
                    {
                         
                        // Set the new sting as
                        // second candidate
                        // for majority
                        t = head.i;
                        q = 1;
                    }
                    else
                    {
                         
                        // Decrease the frequency
                        p = p - 1;
                        q = q - 1;
                    }
                }
            }
        }
        head = head.next;
    }
    head = ptr;
    p = 0;
    q = 0;
 
    // Check the frequency of two
    // final selected candidate linklist
    while (head != null)
    {
        if (s.equals(head.i))
        {
             
            // Increase the frequency
            // of first candidate
            p = 1;
        }
        else
        {
            if (t.equals(head.i))
            {
                 
                // Increase the frequency
                // of second candidate
                q = 1;
            }
        }
        head = head.next;
    }
     
    // Return the String with
    // higher frequency
    if (p > q)
    {
        return s;
    }
    else
    {
        return t;
    }
}
 
// Driver Code
public static void main(String []arg)
{
    node ptr = null;
    node head = newnode("geeks");
    head.next = newnode("geeks");
    head.next.next = newnode("abcd");
    head.next.next.next = newnode("game");
    head.next.next.next.next = newnode("game");
    head.next.next.next.next.next = newnode("knight");
    head.next.next.next.next.next.next = newnode("harry");
    head.next.next.next.next.next.next.next = newnode("geeks");
 
    System.out.println(Majority_in_linklist(head));
}
}
 
// This code is contributed by rutvik_56

Python3

# Python3 program to find an element
# with frequency of at least N / 3
# in a linked list
 
# Structure of a node
# for the linked list
class Node:
    def __init__(self, s):
         
        self.i = s
        self.next = None
         
# Function to find and return
# the element with frequency
# of at least N/3
def Majority_in_linklist(head):
     
    # Candidates for
    # being the required
    # majority element
    s, t = "", ""
     
    # Store the frequencies
    # of the respective candidates
    p, q = 0, 0
    ptr = None
     
    # Iterate all nodes
    while head != None:
        if s == head.i:
             
            # Increase frequency
            # of candidate s
            p = p + 1
        else:
            if t == head.i:
                 
                # Increase frequency
                # of candidate t
                q = q + 1
            else:
                if p == 0:
                     
                    # Set the new sting as
                    # candidate for majority
                    s = head.i
                    p = 1
                else:
                    if q == 0:
                         
                        # Set the new sting as
                        # second candidate
                        # for majority
                        t = head.i
                        q = 1
                    else:
                         
                        # Decrease the frequency
                        p = p - 1
                        q = q - 1
                         
        head = head.next
         
    head = ptr
    p = 0
    q = 0
     
    # Check the frequency of two
    # final selected candidate linklist
    while head != None:
        if s == head.i:
             
            # Increase the frequency
            # of first candidate
            p = 1
        else:
            if t == head.i:
                 
                # Increase the frequency
                # of second candidate
                q = 1
                 
        head = head.next
     
    # Return the string with
    # higher frequency
    if p > q:
        return s
    else:
        return t
 
# Driver code
ptr = None
head = Node("geeks")
head.next = Node("geeks")
head.next.next = Node("abcd")
head.next.next.next = Node("game")
head.next.next.next.next = Node("game")
head.next.next.next.next.next = Node("knight")
head.next.next.next.next.next.next = Node("harry")
head.next.next.next.next.next.next.next = Node("geeks")
 
print(Majority_in_linklist(head))
 
# This code is contributed by stutipathak31jan

C#

// C# program to find an element with
// frequency of at least N / 3 in a
// linked list
using System;
using System.Collections;
using System.Collections.Generic;
 
class GFG{
  
// Structure of a node
// for the linked list
class node
{
    public string i;
    public node next = null;
};
  
// Utility function to
// create a node
static node newnode(string s)
{
    node temp = new node();
    temp.i = s;
    temp.next = null;
    return temp;
}
  
// Function to find and return
// the element with frequency
// of at least N/3
static string Majority_in_linklist(node head)
{
     
    // Candidates for
    // being the required
    // majority element
    string s = "";
    string t = "";
      
    // Store the frequencies
    // of the respective candidates
    int p = 0, q = 0;
    node ptr = null;
  
    // Iterate all nodes
    while (head != null)
    {
        if (s.Equals(head.i))
        {
              
            // Increase frequency
            // of candidate s
            p = p + 1;
        }
        else
        {
            if (t.Equals(head.i))
            {
                  
                // Increase frequency
                // of candidate t
                q = q + 1;
            }
            else
            {
                if (p == 0)
                {
                      
                    // Set the new sting as
                    // candidate for majority
                    s = head.i;
                    p = 1;
                }
                else
                {
                    if (q == 0)
                    {
                          
                        // Set the new sting as
                        // second candidate
                        // for majority
                        t = head.i;
                        q = 1;
                    }
                    else
                    {
                          
                        // Decrease the frequency
                        p = p - 1;
                        q = q - 1;
                    }
                }
            }
        }
        head = head.next;
    }
    head = ptr;
    p = 0;
    q = 0;
  
    // Check the frequency of two
    // final selected candidate linklist
    while (head != null)
    {
        if (s.Equals(head.i))
        {
              
            // Increase the frequency
            // of first candidate
            p = 1;
        }
        else
        {
            if (t.Equals(head.i))
            {
                  
                // Increase the frequency
                // of second candidate
                q = 1;
            }
        }
        head = head.next;
    }
      
    // Return the string with
    // higher frequency
    if (p > q)
    {
        return s;
    }
    else
    {
        return t;
    }
}
  
// Driver Code
public static void Main(string []arg)
{
    node head = newnode("geeks");
    head.next = newnode("geeks");
    head.next.next = newnode("abcd");
    head.next.next.next = newnode("game");
    head.next.next.next.next = newnode("game");
    head.next.next.next.next.next = newnode("knight");
    head.next.next.next.next.next.next = newnode("harry");
    head.next.next.next.next.next.next.next = newnode("geeks");
  
    Console.Write(Majority_in_linklist(head));
}
}
 
// This code is contributed by pratham76

输出：

geeks

时间复杂度： O(N)
辅助空间： O(1)

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