用于在链表中搜索元素的 C++ 程序

编写一个函数，在给定的单链表中搜索给定的键“x”。如果 x 存在于链表中，则该函数应返回 true，否则返回 false。

bool search(Node *head, int x)

例如，如果要搜索的键是 15，链表是 14->21->11->30->10，那么函数应该返回 false。如果要搜索的键是 14，那么函数应该返回 true。
迭代解决方案：

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 

以下是上述算法的迭代实现以搜索给定的键。

C++

// Iterative C++ program to search 
// an element in linked list 
#include 
using namespace std; 
  
// Link list node
class Node 
{ 
    public:
    int key; 
    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(Node** head_ref, int new_key) 
{ 
    // Allocate node
    Node* new_node = new Node();
  
    // Put in the key
    new_node->key = new_key; 
  
    // 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; 
} 
  
// Checks whether the value x is 
// present in linked list 
bool search(Node* head, int x) 
{ 
    Node* current = head; 
    while (current != NULL) 
    { 
        if (current->key == x) 
            return true; 
        current = current->next; 
    } 
    return false; 
} 
  
// Driver code
int main() 
{ 
    // Start with the empty list
    Node* head = NULL; 
    int x = 21; 
  
    // Use push() to construct list 
    // 14->21->11->30->10 
    push(&head, 10); 
    push(&head, 30); 
    push(&head, 11); 
    push(&head, 21); 
    push(&head, 14); 
  
    search(head, 21)? cout<<"Yes" : cout<<"No"; 
    return 0; 
} 
// This is code is contributed by rathbhupendra

输出：

Yes

递归解决方案：

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) 

以下是上述算法的递归实现，用于搜索给定的键。

C++

// Recursive C++ program to search
// an element in linked list 
#include  
using namespace std;
  
// Link list node
struct Node 
{ 
    int key; 
    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_key) 
{ 
    // Allocate node
    struct Node* new_node = 
           (struct Node*) malloc(sizeof(struct Node)); 
  
    // Put in the key
    new_node->key = new_key; 
  
    // 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; 
} 
  
// Checks whether the value x is 
// present in linked list
bool search(struct Node* head, int x) 
{ 
    // Base case 
    if (head == NULL) 
        return false; 
      
    // If key is present in current 
    // node, return true 
    if (head->key == x) 
        return true; 
  
    // Recur for remaining list 
    return search(head->next, x); 
} 
  
// Driver code
int main() 
{ 
    // Start with the empty list
    struct Node* head = NULL; 
    int x = 21; 
  
    // Use push() to construct list 
    // 14->21->11->30->10 
    push(&head, 10); 
    push(&head, 30); 
    push(&head, 11); 
    push(&head, 21); 
    push(&head, 14); 
  
    search(head, 21)? cout << "Yes" : cout << "No"; 
    return 0; 
} 
// This code is contributed by SHUBHAMSINGH10

输出：

Yes

有关详细信息，请参阅有关在链接列表（迭代和递归）中搜索元素的完整文章！