原文：
给定一个大小为n的单链表和另一个键k，我们必须找到单链表中存在键k的概率。
示例：
输入：
list = 2 -> 3 -> 3 -> 3 -> 4 -> 2, key = 5输出：0
说明：
由于列表中不存在值为 5 的键，因此在链表中找到键的概率为 0。
输入：
list = 2 -> 3 -> 5 -> 1 -> 9 -> 8 -> 0 -> 7 -> 6 -> 5, key = 5输出：0.2
方法：
在单链表中找到关键元素k的概率如下：
概率等于元素
k的出现次数/链表的大小
在我们的方法中，我们将首先计算单链表中元素k的数量，然后通过将k的出现次数除以单链表的大小来计算概率。
以下是上述方法的实现：
c
// c code to find the probability
// of finding an element
// in a singly linked list
#include
#include
// link list node
struct node {
int data;
struct node* next;
};
/* 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
struct node* new_node
= (struct node*)malloc(
sizeof(struct 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;
}
// counts nnumber of nodes in linked list
int getcount(struct node* head)
{
// initialize count
int count = 0;
// initialize current
struct node* current = head;
while (current != null) {
count ;
current = current->next;
}
return count;
}
float kpresentprobability(
struct node* head,
int n, int k)
{
// initialize count
float count = 0;
// initialize current
struct node* current = head;
while (current != null) {
if (current->data == k)
count ;
current = current->next;
}
return count / n;
}
// driver code
int main()
{
// start with the empty list
struct node* head = null;
// use push() to construct below list
// 1->2->1->3->1
push(&head, 2);
push(&head, 3);
push(&head, 5);
push(&head, 1);
push(&head, 9);
push(&head, 8);
push(&head, 0);
push(&head, 7);
push(&head, 6);
push(&head, 5);
printf("%.1f",
kpresentprobability(
head, getcount(head), 5));
return 0;
}
java
// java code to find the probability
// of finding an element
// in a singly linked list
class gfg{
// link list node
static class node
{
int data;
node next;
};
// 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;
}
// counts nnumber of nodes
// in linked list
static int getcount(node head)
{
// initialize count
int count = 0;
// initialize current
node current = head;
while (current != null)
{
count ;
current = current.next;
}
return count;
}
static float kpresentprobability(node head,
int n, int k)
{
// initialize count
float count = 0;
// initialize current
node current = head;
while (current != null)
{
if (current.data == k)
count ;
current = current.next;
}
return count / n;
}
// driver code
public static void main(string[] args)
{
// start with the empty list
node head = null;
// use push() to conbelow list
// 1.2.1.3.1
head = push(head, 2);
head = push(head, 3);
head = push(head, 5);
head = push(head, 1);
head = push(head, 9);
head = push(head, 8);
head = push(head, 0);
head = push(head, 7);
head = push(head, 6);
head = push(head, 5);
system.out.printf("%.1f", kpresentprobability(
head, getcount(head), 5));
}
}
// this code is contributed by shikhasingrajput
python3
# python3 code to find the probability
# of finding an element
# in a singly linked list
# node class
class node:
def __init__(self, data, next = none):
self.data = data
self.next = none
class linkedlist:
def __init__(self):
self.head = none
def push(self, data):
# allocate the node &
# put the data
new_node = node(data)
# make the next of new node as head
new_node.next = self.head
# move the head to point to new node
self.head = new_node
# counts the number of nodes in linkedlist
def getcount(self):
# initialize current
current = self.head
# initialize count
count = 0
while current is not none:
count = 1
current = current.next
return count
def kpresentprobability(self, n, k):
# initialize current
current = self.head
# initialize count
count = 0.0
while current is not none:
if current.data == k:
count = 1
current = current.next
return count / n
# driver code
if __name__ == "__main__":
# start with empty list
llist = linkedlist()
# use push to construct the linked list
llist.push(2)
llist.push(3)
llist.push(5)
llist.push(1)
llist.push(9)
llist.push(8)
llist.push(0)
llist.push(7)
llist.push(6)
llist.push(5)
print(llist.kpresentprobability(
llist.getcount(), 5))
# this code is contributed by kevalshah5
c
// c# code to find the probability
// of finding an element
// in a singly linked list
using system;
class gfg{
// link list node
class node
{
public int data;
public node next;
};
// 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;
}
// counts nnumber of nodes
// in linked list
static int getcount(node head)
{
// initialize count
int count = 0;
// initialize current
node current = head;
while (current != null)
{
count ;
current = current.next;
}
return count;
}
static float kpresentprobability(node head,
int n, int k)
{
// initialize count
float count = 0;
// initialize current
node current = head;
while (current != null)
{
if (current.data == k)
count ;
current = current.next;
}
return count / n;
}
// driver code
public static void main(string[] args)
{
// start with the empty list
node head = null;
// use push() to conbelow list
// 1.2.1.3.1
head = push(head, 2);
head = push(head, 3);
head = push(head, 5);
head = push(head, 1);
head = push(head, 9);
head = push(head, 8);
head = push(head, 0);
head = push(head, 7);
head = push(head, 6);
head = push(head, 5);
console.write("{0:f1}", kpresentprobability(
head, getcount(head), 5));
}
}
// this code is contributed by princi singh
输出：
0.2
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