数据结构4：基于单链表的通讯录项目-CFANZ编程社区

struct ListNode* middleNode(struct ListNode* head) {
    struct ListNode* slow=head;
    struct ListNode* fast=head;
    while(fast!=NULL&&fast->next!=NULL)
    {
        fast=fast->next->next;
        slow=slow->next;
    }
    return slow;
}

1.3 收获

快慢指针的用法：
链表长度为奇数：当快指针走到尾节点时，慢指针正好走到中间节点。
链表长度为偶数：当快指针走到NULL时，慢指针正好走到第二个中间节点。

二、移除链表元素

2.1 题目

2.2 题解

//创建虚拟节点法
struct ListNode* removeElements1(struct ListNode* head, int val) {
    struct ListNode* dummyHead = malloc(sizeof(struct ListNode));
    dummyHead->next = head;
    struct ListNode* temp = dummyHead;
    while (temp->next != NULL) {
        if (temp->next->val == val) {
            temp->next = temp->next->next;
        }
        else {
            temp = temp->next;
        }
    }
    return dummyHead->next;
}
//创建新链表法
struct ListNode* removeElements3(struct ListNode* head, int val) {
    struct ListNode* phead = NULL;
    struct ListNode* ptail = NULL;

    struct ListNode* pcur = head;
    while (pcur) {
        if (pcur->val != val) {
            if (phead == NULL) {
                phead = ptail = pcur;
            }
            else {
                ptail->next = pcur;
                ptail = ptail->next;
            }
        }
        pcur = pcur->next;
    }
    if (phead) {
        ptail->next = NULL;
    }
    return phead;
}
//双指针法
struct ListNode* removeElements(struct ListNode* head, int val) {
    while (NULL != head && head->val == val) {
        head = head->next;
    }
    struct ListNode* pcur = head;
    struct ListNode* prev = head;
    while (pcur)
    {
        if (pcur->val != val)
        {
            prev = pcur;
        }
        else
        {
            prev->next = pcur->next;
        }
        pcur = pcur->next;
    }
    return head;
}
//递归做法
struct ListNode* removeElements2(struct ListNode* head, int val) {
    if (head == NULL) {
        return head;
    }
    head->next = removeElements(head->next, val);
    if (head->val == val)
    {
        return head->next;
    }
    else
    {
        return head;
    }
}

2.3 收获

创建虚拟头节点：好处是可以只维护一个指针
创建新链表法：思路简单：仅仅为挑选符合条件的数据复制即可
双指针法：保存前一个链表的指针

2.4递归详解

停止条件：当遍历链表的指针为空时。
如何递归：判断节点值是否等于给定值，并决定是否要删除。

三、反转链表

3.1 题目

3.2 题解

struct ListNode* reverseList(struct ListNode* head) {
    if (head == NULL) {
        return head;
    }
    struct ListNode* a = NULL;
    struct ListNode* b = head;
    struct ListNode* c = head->next;

    while (b) {
        b->next = a;
        a = b;
        b = c;
        if (c) {
            c = c->next;
        }
    }
    return a;
}

3.3 解释

三指针法：循环保留三个指针，修改朝向即可；

四、合并两个有序列表

4.1 题目

4.2 题解

 //创建新链表法
struct ListNode* mergeTwoLists(struct ListNode* list1, struct ListNode* list2) {
    if (list1 == NULL)
    {
        return list2;
    }
    if (list2 == NULL)
    {
        return list1;
    }
    if (list1 == NULL && list2 == NULL)
    {
        return NULL;
    }
    struct ListNode* l1 = list1;
    struct ListNode* l2 = list2;
    struct ListNode* newhead;
    struct ListNode* newtail;
    newhead = newtail = NULL;
    while (l1 && l2)
    {
        if (l1->val < l2->val)
        {
            if (newhead == NULL)
            {
                newhead = newtail = l1;
            }
            else
            {
                newtail->next = l1;
                newtail = newtail->next;
            }
            l1 = l1->next;
        }
        else
        {
            if (newhead == NULL)
            {
                newhead = newtail = l2;
            }
            else
            {
                newtail->next = l2;
                newtail = newtail->next;
            }
            l2 = l2->next;
        }
    }
    if (l1 == NULL && l2 != NULL)
    {
        newtail->next = l2;
        while (l2->next)
        {
            l2 = l2->next;
        }
        newtail = l2;
    }
    if (l2 == NULL && l1 != NULL)
    {
        newtail->next = l1;
        while (l1->next)
        {
            l1 = l1->next;
        }
        newtail = l1;
    }
    return newhead;
}
//递归
struct ListNode* mergeTwoLists(struct ListNode* list1, struct ListNode* list2) {
    struct ListNode* l1 = list1;
    struct ListNode* l2 = list2;
    if (l1 == NULL) {
        return l2;
    }
    if (l2 == NULL) {
        return l1;
    }
    if (list1 == NULL && list2 == NULL)
    {
        return NULL;
    }
    if (l1->val <= l2->val) {
        l1->next = mergeTwoLists(l1->next, l2);
        return l1;
    }
    l2->next = mergeTwoLists(l1, l2->next);
    return l2;
}