插入节点
在链表中插入节点有以下三种情形：
1、在链表头部
2、在特定节点
3、在链表尾部
1）在头部添加（4步操作）
新节点添加到头部，将成为新的头节点，以下将节点添加到链表头部的函数是push，push接收指向链表头部的指针，然后将指针修改
指向新的节点：
c语言：
/* Given a reference (pointer to pointer) to the head of a list
and an int, inserts a new node on the front of the list. */
void push(struct Node** head_ref, int new_data)
{
/* 1. allocate node */
struct Node* new_node = (struct Node*) malloc(sizeof(struct Node)); /* 2. put in the data */ new_node->data = new_data; /* 3. Make next of new node as head */ new_node->next = (*head_ref); /* 4. move the head to point to the new node */ (*head_ref) = new_node; }
Java：
/* This function is in LinkedList class. Inserts a
new Node at front of the list. This method is
defined inside LinkedList class shown above */
public void push(int new_data)
{
/* 1 & 2: Allocate the Node &
Put in the data*/ Node new_node = new Node(new_data); /* 3. Make next of new Node as head */ new_node.next = head; /* 4. Move the head to point to new Node */ head = new_node; }
push时间复杂度O(1)
2）在特定节点（5步操作）
c语言：
/* Given a node prev_node, insert a new node after the given
prev_node */
void insertAfter(struct Node* prev_node, int new_data)
{
/*1. check if the given prev_node is NULL */
if (prev_node == NULL) { printf("the given previous node cannot be NULL"); return; } /* 2. allocate new node */ struct Node* new_node =(struct Node*) malloc(sizeof(struct Node)); /* 3. put in the data */ new_node->data = new_data; /* 4. Make next of new node as next of prev_node */ new_node->next = prev_node->next; /* 5. move the next of prev_node as new_node */ prev_node->next = new_node; }
java：
/* This function is in LinkedList class.
Inserts a new node after the given prev_node. This method is
defined inside LinkedList class shown above */
public void insertAfter(Node prev_node, int new_data)
{
/* 1. Check if the given Node is null */
if (prev_node == null) { System.out.println("The given previous node cannot be null"); return; } /* 2. Allocate the Node & 3. Put in the data*/ Node new_node = new Node(new_data); /* 4. Make next of new Node as next of prev_node */ new_node.next = prev_node.next; /* 5. make next of prev_node as new_node */ prev_node.next = new_node; }
因为已经给定了节点，所以insertAfter时间复杂为O(1)
3）在链表尾部（6个步骤）
我们需要遍历链表，得到最后一个节点，然后把新节点追加到最后，然后让他成为新的尾部节点
c语言：
/* Given a reference (pointer to pointer) to the head
of a list and an int, appends a new node at the end */
void append(struct Node** head_ref, int new_data)
{
/* 1. allocate node */
struct Node* new_node = (struct Node*) malloc(sizeof(struct Node)); struct Node *last = *head_ref; /* used in step 5*/ /* 2. put in the data */ new_node->data = new_data; /* 3. This new node is going to be the last node, so make next of it as NULL*/ new_node->next = NULL; /* 4. If the Linked List is empty, then make the new node as head */ if (*head_ref == NULL) { *head_ref = new_node; return; } /* 5. Else traverse till the last node */ while (last->next != NULL) last = last->next; /* 6. Change the next of last node */ last->next = new_node; return; }
java：
/* Appends a new node at the end. This method is
defined inside LinkedList class shown above */
public void append(int new_data)
{
/* 1. Allocate the Node &
2. Put in the data
3. Set next as null */ Node new_node = new Node(new_data); /* 4. If the Linked List is empty, then make the new node as head */ if (head == null) { head = new Node(new_data); return; } /* 4. This new node is going to be the last node, so make next of it as null */ new_node.next = null; /* 5. Else traverse till the last node */ Node last = head; while (last.next != null) last = last.next; /* 6. Change the next of last node */ last.next = new_node; return; }
因为需要从头到尾循环，所以append时间复杂度为O(n)，其中n是节点的个数。
当然你可以修改结构，保存尾节点，使其复杂度降为O(1)
来源：https://www.geeksforgeeks.org/linked-list-set-2-inserting-a-node/