c - 链表返回单链表 C 中的根节点
问题描述
我正在编写一个程序来删除大于 x 的节点。我想删除后返回链表的根节点。我将新的根节点保存在变量“root”中。但不幸的是,当我在 main 函数中获取 root 时,该值似乎为 0。任何人都可以帮我确定问题吗?**
// structure of a node
struct Node {
int data;
Node* next;
};
// function to get a new node
Node* getNode(int data)
{
Node* newNode = (Node*)malloc(sizeof(struct 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
Node *deleteGreaterNodes(Node* head_ref, int x)
{
Node *temp = head_ref, *prev;
static Node *root;
// 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
}
root = 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 0;
// Unlink the node from linked list
prev->next = temp->next;
free (temp); // Free memory
// Update Temp for next iteration of
// outer loop
temp = prev->next;
}
return head_ref;
}
// function to a print a linked list
void printList(Node* head)
{
while (head) {
printf("%d ",head->data);
head = head->next;
}
}
// Driver program to test above
int main()
{
static Node *root;
// 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 = 3;
printf("Original List: ");
printList(head);
root = deleteGreaterNodes(head, x);
printf("\nModified List: ");
printList(root);
return 0;
}
解决方案
一种简单的实现方式deleteGreaterNodes
是使用额外的间接级别来访问Node *
链接。
// function to delete all the nodes from the list
// that are greater than the specified value x
Node *deleteGreaterNodes(Node* head_ref, int x)
{
Node **link = &head_ref; // Point to the pointer to the first node.
Node *cur;
while ((cur = *link) != NULL) {
if (cur->data > x) {
// The current node needs to be deleted from the list.
// Change the link pointing to the current node to point to the next node.
*link = cur->next;
// Delete the current node.
free(cur);
} else {
// The current node is to be kept on the list.
// Get a pointer to the link to the next node.
link = &cur->next;
}
}
return head_ref;
}
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