用于从排序链表中删除所有重复项的 C++ 程序

给定一个排序的链表，删除所有具有重复数字（所有出现）的节点，只留下在原始列表中出现一次的数字。
例子：

Input: 23->28->28->35->49->49->53->53
Output: 23->35

Input: 11->11->11->11->75->75
Output: empty List

请注意，这与从链接列表中删除重复项不同

这个想法是维护一个指向节点的指针（prev） ，该节点恰好位于我们正在检查重复的节点块之前。在第一个示例中，当我们检查节点 28 的重复项时，指针prev将指向 23。一旦我们到达值为 28 的最后一个重复节点（将其命名为当前指针），我们可以使 prev 节点的下一个字段成为当前的下一个并更新current=current.next 。这将删除具有重复项的值为 28 的节点块。

C++

// C++ program to remove all occurrences 
// of duplicates from a sorted linked list.
#include 
using namespace std;
  
// A linked list node
struct Node
{
    int data;
    struct Node *next;
};
  
// Utility function 
// to create a new Node
struct Node *newNode(int data)
{
    Node *temp = new Node;
    temp -> data = data;
    temp -> next = NULL;
    return temp;
}
  
// Function to print nodes 
// in a given linked list.
void printList(struct Node *node)
{
    while (node != NULL)
    {
        printf("%d ", node -> data);
        node = node -> next;
    }
}
  
// Function to remove all occurrences
// of duplicate elements
void removeAllDuplicates(struct Node* &start)
{
    // Create a dummy node that acts like 
    // a fake head of list pointing to the 
    // original head
    Node* dummy = new Node;
  
    // Dummy node points to the original head
    dummy -> next = start;
      
    // Node pointing to last node which has 
    // no duplicate.
    Node* prev = dummy;
      
    // Node used to traverse the linked list.
    Node* current = start;
  
    while(current != NULL)
    {
        // Until the current and previous 
        // values are same, keep updating current
        while(current -> next != NULL &&
              prev -> next -> data == current -> next -> data)
            current = current -> next;
  
        // if current has unique value 
        // i.e current is not updated, 
        // Move the prev pointer to 
        // next node
        if (prev -> next == current)
            prev = prev -> next;
  
        // When current is updated to the 
        // last duplicate value of that segment, 
        // make prev the next of current
        else
            prev -> next = current -> next;
  
        current = current -> next;
    }
  
    // Update original head to the next 
    // of dummy node 
    start = dummy -> next;
}
  
// Driver Code
int main() 
{
    // 23->28->28->35->49->49->53->53
    struct Node* start = newNode(23);
    start -> next = newNode(28);
    start -> next -> next = newNode(28);
    start -> next -> 
             next -> next = newNode(35);
    start -> next -> 
    next -> next -> next = newNode(49);
    start -> next -> 
    next -> next -> 
    next -> next = newNode(49);
    start -> next -> 
    next -> next -> 
    next -> next -> next = newNode(53);
    start -> next -> 
    next -> next -> 
    next -> next -> 
    next -> next = newNode(53);
    cout << "List before removal " <<
            "of duplicates";
    printList(start);
      
    removeAllDuplicates(start);
      
    cout << "List after removal " << 
            "of duplicates";
    printList(start);
    return 0;
}
// This code is contributed by NIKHIL JINDAL

输出：

List before removal of duplicates
23 28 28 35 49 49 53 53 
List after removal of duplicates
23 35 

时间复杂度： O(n)
有关详细信息，请参阅有关从排序的链接列表中删除所有重复项的完整文章！