从列表中删除所有大于 x 的节点

给定一个链表，问题是从链表中删除所有大于指定值x的节点。

例子：

Input : list: 7->3->4->8->5->1
        x = 6
Output : 3->4->5->1

Input : list: 1->8->7->3->7->10
        x = 7
Output : 1->7->3->7

来源：微软面试体验 |设置 169。

方法：这主要是删除给定键的第一次出现的帖子的变体。

我们需要首先检查头节点上所有大于“x”的出现，删除它们并适当地更改头节点。然后我们需要检查循环中的所有事件并一一删除它们。

// C++ implementation to delete all the nodes from the list
// that are greater than the specified value x
#include 
  
using namespace std;
  
// structure of a node
struct Node {
    int data;
    Node* next;
};
  
// function to get a new node
Node* getNode(int data)
{
    Node* newNode = new Node;
    newNode->data = data;
    newNode->next = NULL;
    return newNode;
}
  
// function to delete all the nodes from the list
// that are greater than the specified value x
void deleteGreaterNodes(Node** head_ref, int x)
{
    Node *temp = *head_ref, *prev;
  
    // If head node itself holds the value greater than 'x'
    if (temp != NULL && temp->data > x) {
        *head_ref = temp->next; // Changed head
        free(temp); // free old head
        temp = *head_ref; // Change temp
    }
  
    // Delete occurrences other than head
    while (temp != NULL) {
  
        // Search for the node to be deleted, 
        // keep track of the previous node as we 
        // need to change 'prev->next'
        while (temp != NULL && temp->data <= x) {
            prev = temp;
            temp = temp->next;
        }
  
        // If required value node was not present
        // in linked list
        if (temp == NULL)
            return;
  
        // Unlink the node from linked list
        prev->next = temp->next;
  
        delete temp; // Free memory
  
        // Update Temp for next iteration of 
        // outer loop
        temp = prev->next;
    }
}
  
// function to a print a linked list
void printList(Node* head)
{
    while (head) {
        cout << head->data << " ";
        head = head->next;
    }
}
  
// Driver program to test above
int main()
{
    // Create list: 7->3->4->8->5->1
    Node* head = getNode(7);
    head->next = getNode(3);
    head->next->next = getNode(4);
    head->next->next->next = getNode(8);
    head->next->next->next->next = getNode(5);
    head->next->next->next->next->next = getNode(1);
  
    int x = 6;
  
    cout << "Original List: ";
    printList(head);
  
    deleteGreaterNodes(&head, x);
  
    cout << "\nModified List: ";
    printList(head);
  
    return 0;
}

输出：

Original List: 7 3 4 8 5 1 
Modified List: 3 4 5 1

时间复杂度：O(n)。

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