用于在链表中搜索元素的 Javascript 程序
原文:https://www . geesforgeks . org/JavaScript-program-for-search-in-a-element-in-a-link-list/
编写一个函数，在给定的单链表中搜索给定的键“x”。如果 x 出现在链表中，函数应该返回 true，否则返回 false。
bool search(Node *head, int x)
例如，如果要搜索的关键字是 15，链表是 14->21->11->30->10，那么函数应该返回 false。如果要搜索的关键字是 14，那么函数应该返回 true。 迭代求解:
1) Initialize a node pointer, current = head.
2) Do following while current is not NULL
a) current->key is equal to the key being searched return true.
b) current = current->next
3) Return false
下面是上述算法的迭代实现，以搜索给定的关键字。
java 描述语言
<script>
// Iterative javascript program
// to search an element
// in linked list
//Node class
class Node
{
constructor(d)
{
this.data = d;
this.next = null;
}
}
// Linked list class
// Head of list
var head;
// Inserts a new node at the front of the list
function push(new_data)
{
// Allocate new node and putting data
var new_node = new Node(new_data);
// Make next of new node as head
new_node.next = head;
// Move the head to point to new Node
head = new_node;
}
// Checks whether the value
// x is present in linked list
function search(head , x)
{
// Initialize current
var current = head;
while (current != null)
{
if (current.data == x)
// Data found
return true;
current = current.next;
}
// Data not found
return false;
}
// Driver code
// Start with the empty list
// Use push() to construct list
// 14->21->11->30->10
push(10);
push(30);
push(11);
push(21);
push(14);
if (search(head, 21))
document.write("Yes");
else
document.write("No");
// This code contributed by aashish1995
</script>
输出:
Yes
递归解:
bool search(head, x)
1) If head is NULL, return false.
2) If head's key is same as x, return true;
3) Else return search(head->next, x)
下面是上述算法的递归实现，用于搜索给定的关键字。
java 描述语言
<script>
// Recursive javascript program to search
// an element in linked list
// Node class
class Node
{
constructor(val)
{
this.data = val;
this.next = null;
}
}
// Linked list class
// Head of list
var head;
// Inserts a new node at the front
// of the list
function push(new_data)
{
// Allocate new node and putting data
var new_node = new Node(new_data);
// Make next of new node as head
new_node.next = head;
// Move the head to point to new Node
head = new_node;
}
// Checks whether the value x is present
// in linked list
function search(head, x)
{
// Base case
if (head == null)
return false;
// If key is present in current node,
// return true
if (head.data == x)
return true;
// Recur for remaining list
return search(head.next, x);
}
// Driver code
// Start with the empty list
// Use push() to construct list
// 14->21->11->30->10
push(10);
push(30);
push(11);
push(21);
push(14);
if (search(head, 21))
document.write("Yes");
else
document.write("No");
// This code contributed by gauravrajput1
</script>
输出:
Yes
更多细节请参考完整的文章在链表中搜索元素(迭代和递归)！
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