用于将元素的所有出现移动到链表中结束的 C++ 程序

给定一个链表和其中的一个键，任务是将所有出现的给定键移动到链表的末尾，保持所有其他元素的顺序相同。

例子：

Input  : 1 -> 2 -> 2 -> 4 -> 3
         key = 2 
Output : 1 -> 4 -> 3 -> 2 -> 2

Input  : 6 -> 6 -> 7 -> 6 -> 3 -> 10
         key = 6
Output : 7 -> 3 -> 10 -> 6 -> 6 -> 6

一个简单的解决方案是在链表中逐一查找给定键的所有出现。对于每个找到的事件，将其插入到末尾。我们这样做直到给定键的所有出现都移到最后。

时间复杂度： O(n ² )

有效的解决方案1：是保留两个指针：
pCrawl => 一个一个遍历整个列表的指针。
pKey => 如果找到键，则指向该键的出现。其他与 pCrawl 相同。
我们从链表的头部开始上述两个指针。只有当pKey不指向某个键时，我们才移动pKey 。我们总是移动pCrawl 。所以，当pCrawl和pKey不相同时，我们肯定已经找到了一个位于pCrawl之前的 key，所以我们在pCrawl和pKey之间交换，并将pKey移动到下一个位置。循环不变量是，在交换数据之后，从pKey到pCrawl的所有元素都是键。

下面是这种方法的实现。

C++

// C++ program to move all occurrences of a
// given key to end.
#include 
  
// A Linked list Node
struct Node {
    int data;
    struct Node* next;
};
  
// A utility function to create a new node.
struct Node* newNode(int x)
{
    Node* temp = new Node;
    temp->data = x;
    temp->next = NULL;
}
  
// Utility function to print the elements
// in Linked list
void printList(Node* head)
{
    struct Node* temp = head;
    while (temp != NULL) {
        printf("%d ", temp->data);
        temp = temp->next;
    }
    printf("
");
}
  
// Moves all occurrences of given key to
// end of linked list.
void moveToEnd(Node* head, int key)
{
    // Keeps track of locations where key
    // is present.
    struct Node* pKey = head;
  
    // Traverse list
    struct Node* pCrawl = head;
    while (pCrawl != NULL) {
        // If current pointer is not same as pointer
        // to a key location, then we must have found
        // a key in linked list. We swap data of pCrawl
        // and pKey and move pKey to next position.
        if (pCrawl != pKey && pCrawl->data != key) {
            pKey->data = pCrawl->data;
            pCrawl->data = key;
            pKey = pKey->next;
        }
  
        // Find next position where key is present
        if (pKey->data != key)
            pKey = pKey->next;
  
        // Moving to next Node
        pCrawl = pCrawl->next;
    }
}
  
// Driver code
int main()
{
    Node* head = newNode(10);
    head->next = newNode(20);
    head->next->next = newNode(10);
    head->next->next->next = newNode(30);
    head->next->next->next->next = newNode(40);
    head->next->next->next->next->next = newNode(10);
    head->next->next->next->next->next->next = newNode(60);
  
    printf("Before moveToEnd(), the Linked list is
");
    printList(head);
  
    int key = 10;
    moveToEnd(head, key);
  
    printf("
After moveToEnd(), the Linked list is
");
    printList(head);
  
    return 0;
}

输出：

Before moveToEnd(), the Linked list is
10 20 10 30 40 10 60 

After moveToEnd(), the Linked list is
20 30 40 60 10 10 10

时间复杂度： O(n) 只需要遍历一次列表。

高效解决方案 2：
1. 遍历链表，在尾部取一个指针。
2. 现在，检查密钥和节点->数据。如果它们相等，则将节点移动到 last-next，否则继续前进。

C++

// C++ code to remove key element to end of linked list
#include
using namespace std;
  
// A Linked list Node
struct Node 
{
    int data;
    struct Node* next;
};
  
// A utility function to create a new node.
struct Node* newNode(int x)
{
    Node* temp = new Node;
    temp->data = x;
    temp->next = NULL;
}
  
// Function to remove key to end
Node *keyToEnd(Node* head, int key)
{
  
    // Node to keep pointing to tail
    Node* tail = head;
    if (head == NULL) 
    {
        return NULL;
    }
    while (tail->next != NULL) 
    {
        tail = tail->next;
    }
      
    // Node to point to last of linked list
    Node* last = tail;
    Node* current = head;
    Node* prev = NULL;
      
    // Node prev2 to point to previous when head.data!=key
    Node* prev2 = NULL;
      
    // loop to perform operations to remove key to end
    while (current != tail) 
    {
        if (current->data == key && prev2 == NULL) 
        {
            prev = current;
            current = current->next;
            head = current;
            last->next = prev;
            last = last->next;
            last->next = NULL;
            prev = NULL;
        }
        else
        {
            if (current->data == key && prev2 != NULL)
            {
                prev = current;
                current = current->next;
                prev2->next = current;
                last->next = prev;
                last = last->next;
                last->next = NULL;
            }
            else if (current != tail) 
            {
                prev2 = current;
                current = current->next;
            }
        }
    }
    return head;
}
  
// Function to display linked list
void printList(Node* head) 
{
    struct Node* temp = head;
    while (temp != NULL) 
    {
        printf("%d ", temp->data);
        temp = temp->next;
    }
    printf("\n");
}
  
  
// Driver Code
int main()
{
    Node* root = newNode(5);
    root->next = newNode(2);
    root->next->next = newNode(2);
    root->next->next->next = newNode(7);
    root->next->next->next->next = newNode(2);
    root->next->next->next->next->next = newNode(2);
    root->next->next->next->next->next->next = newNode(2);
  
    int key = 2;
    cout << "Linked List before operations :";
    printList(root);
    cout << "\nLinked List after operations :";
    root = keyToEnd(root, key);
    printList(root);
    return 0;
}
  
// This code is contributed by Rajout-Ji

输出：

Linked List before operations :
5 2 2 7 2 2 2 
Linked List after operations :
5 7 2 2 2 2 2

感谢Ravinder Kumar提出这种方法。

高效的解决方案3：是维护一个单独的键列表。我们将此键列表初始化为空。我们遍历给定的列表。对于找到的每个键，我们将其从原始列表中删除并将其插入到单独的键列表中。我们最终将键列表链接到剩余给定列表的末尾。该解决方案的时间复杂度也是 O(n)，而且它也只需要遍历一次列表。

有关详细信息，请参阅有关将所有出现的元素移动到链接列表中的完整文章！