Singly linked list example

.Basic idea: 1 ) Determine whether the linked list is empty, if it is empty, return NULL

2 ) Traverse the linked list, find the node before the corresponding val , and link it with the next node of val .

3 ) Pit point: there will be a situation where the head node is val , so we need to deal with the problem of the head node additionally.

It is more convenient to use two pointers for this question.

draw

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */


struct ListNode* removeElements(struct ListNode* head, int val){
    while(head&&head->val==val)//找新的头节点，改新头节点的head->val!=val
    {
        head = head->next;//更新head
    }
    if(head==NULL)//判断新头节点是否出现示例2或示例3的极端情况
    {
        return NULL;
    }
    struct ListNode* prev = head;//创建指针，记录当前head的地址
    while(prev->next)
    {
        if(prev->next->val == val)//判断节点val的值是否与val相等。
        {
            prev->next = prev->next->next;//相等就进行2）中的链接
        }
        else 
        {
            prev = prev->next;//更新prev
        }
    }
    return head;
}

two:

reverse linked list

Basic idea: 1) Determine whether the linked list is empty, if it is empty, return NULL

2) Processing method:

i build a new head node and tail-interpolate the original data

ii Change the pointing of the original arrow

Drawing:

i。

Note: The loop end condition is cur==NULL

Every time you need to update cur and newHead, and finally return newHead

Since the loop ends when cur is empty, the next will become a null pointer, which requires special treatment

Code:

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */


struct ListNode* reverseList(struct ListNode* head){
    if(head == NULL)//判断链表是否为空
    {
        return NULL;
    }
    struct ListNode* newHead = NULL,*cur = head,*next = head->next;
    while(cur)
    {
        cur->next = newHead;//头插节点
        newHead = cur;//更新节点newHead
        cur = next;//更新cucr指针
        if(next!=NULL)//防止next成为野指针
        next = next->next;//更新next指针
    }
    return newHead;
}

ii. Forcibly reverse the arrow

Note: 1) For the convenience of processing, three pointer variables are set: prev, cur, next, which respectively record the nodes before, after, and after cur.

2) Similar to the idea of the above figure, it is still linking and updating nodes

Drawing:

Code:

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */


struct ListNode* reverseList(struct ListNode* head){
    if(head==NULL)//判断链表是否为空
        return NULL;
    struct ListNode* prev = NULL;//
    struct ListNode* cur = head;//
    struct ListNode* next = head->next;//创建指针变量

    while(cur)//循环
    {
        cur->next = prev;//链接节点
        prev = cur;//更新prev
        cur = next;//更新cur
        if(next)//防止next成为野指针
            next = next->next;//更新next
    }
    return prev;
}

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