L6.【LeetCode笔记】合并两个有序链表

1.题目

https://leetcode.cn/problems/merge-two-sorted-lists/

将两个升序链表合并为一个新的升序链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。

示例 1：
复制代码
输入：l1 = [1,2,4], l2 = [1,3,4]
输出：[1,1,2,3,4,4]
示例 2：
复制代码
输入：l1 = [], l2 = []
输出：[]
示例 3：
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输入：l1 = [], l2 = [0]
输出：[0]
提示：

两个链表的节点数目范围是 [0, 50]

-100 <= Node.val <= 100

l1 和 l2 均按 非递减顺序 排列

代码模版
cpp 复制代码
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* mergeTwoLists(struct ListNode* list1, struct ListNode* list2) 
{
}

2.自解

一个容易想到的解法:取小的尾插(开一个新的链表)

对于链表list1和list2,可以另外开一个新的链表,再将list1和list2的val复制进新链表的节点,最后返回新链表的头结点的地址即可

不加思索写出以下代码:

cpp 复制代码

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* mergeTwoLists(struct ListNode* list1, struct ListNode* list2) 
{
    struct ListNode* cur1=list1;
    struct ListNode* cur2=list2;

    if (list1==NULL)
        return list2;
    if (list2==NULL)
        return list1;
    
    struct newListNode
    {
        int new_val;
        struct ListNode* new_next;
    };
    struct newListNode* new_next=NULL;
    struct newListNode* newhead=NULL;
    struct newListNode* m_m_new=(struct newListNode*)malloc(sizeof(struct newListNode));
    newhead=m_m_new;
    newhead->new_next=NULL;
    struct newListNode* new_cur=newhead;
    while(cur1!=NULL && cur2!=NULL)
    {
        if (cur1==NULL)
        {
            new_cur->new_val=cur2->val;
            cur2=cur2->next;
            //分配新结点的空间
            struct newListNode* m_new=(struct newListNode*)malloc(sizeof(struct newListNode));
            new_cur->new_next=m_new;
            new_cur=m_new;
            continue;
        }
        if (cur2==NULL)
        {
            new_cur->new_val=cur1->val;
            cur1=cur1->next;
            struct newListNode* m_new=(struct newListNode*)malloc(sizeof(struct newListNode));
            new_cur->new_next=m_new;
            new_cur=m_new;
            continue;
        }
        if (cur1->val<=cur2->val)
        {
            new_cur->new_val=cur1->val;
            cur1=cur1->next;
            struct newListNode* m_new=(struct newListNode*)malloc(sizeof(struct newListNode));
            new_cur->new_next=m_new;
            new_cur=m_new;
        }
        else
        {
            new_cur->new_val=cur2->val;
            cur2=cur2->next;
            //分配新结点的空间
            struct newListNode* m_new=(struct newListNode*)malloc(sizeof(struct newListNode));
            new_cur->new_next=m_new;
            new_cur=m_new;
        }
    }
    new_cur->new_next=NULL;
    new_cur=NULL;
    return newhead;
}

运行时出现问题

发现while循环的条件写错了!!

应该改成

cpp 复制代码

while(!(cur1==NULL && cur2==NULL))

完整代码

cpp 复制代码

struct ListNode* mergeTwoLists(struct ListNode* list1, struct ListNode* list2) 
{
    struct ListNode* cur1=list1;
    struct ListNode* cur2=list2;

    if (list1==NULL)
        return list2;
    if (list2==NULL)
        return list1;
    
    struct newListNode
    {
        int new_val;
        struct ListNode* new_next;
    };
    struct newListNode* new_next=NULL;
    struct newListNode* newhead=NULL;
    struct newListNode* m_m_new=(struct newListNode*)malloc(sizeof(struct newListNode));
    newhead=m_m_new;
    newhead->new_next=NULL;
    struct newListNode* new_cur=newhead;
    struct newListNode* before_new_cur=NULL;

    while(!(cur1==NULL && cur2==NULL))
    {
        if (cur1==NULL)
        {
            new_cur->new_val=cur2->val;
            cur2=cur2->next;
            struct newListNode* m_new=(struct newListNode*)malloc(sizeof(struct newListNode));
            new_cur->new_next=m_new;
            before_new_cur=new_cur;
            new_cur=m_new;
            new_cur->new_next=NULL;
            continue;
        }
        if (cur2==NULL)
        {
            new_cur->new_val=cur1->val;
            cur1=cur1->next;
            struct newListNode* m_new=(struct newListNode*)malloc(sizeof(struct newListNode));
            new_cur->new_next=m_new;
            before_new_cur=new_cur;
            new_cur=m_new;
            continue;
        }
        
        if (cur1->val<=cur2->val)
        {
            new_cur->new_val=cur1->val;
            cur1=cur1->next;
            struct newListNode* m_new=(struct newListNode*)malloc(sizeof(struct newListNode));
            new_cur->new_next=m_new;
            new_cur=m_new;
        }
        else
        {
            new_cur->new_val=cur2->val;
            cur2=cur2->next;           
            struct newListNode* m_new=(struct newListNode*)malloc(sizeof(struct newListNode));
            new_cur->new_next=m_new;
            new_cur=m_new;
        }
    }
   before_new_cur->new_next=NULL;
   return newhead;
}

before_new_cur是当cur1===NULL或cur2==NULL,备份new_cur的前一个节点的地址

提交结果

3.其他解法

方法1:取小的尾插(不开新链表)

cpp 复制代码

struct ListNode* mergeTwoLists(struct ListNode* list1, struct ListNode* list2) 
{
    struct ListNode* cur1=list1;
    struct ListNode* cur2=list2;
    struct ListNode* head=NULL;
    struct ListNode* tail=NULL;

    if (list1==NULL)
        return list2;
    if (list2==NULL)
        return list1;

    while (cur1 && cur2)
    {
        if (cur1->val<cur2->val)
        {
            if (head==NULL)
            {
                head=tail=cur1;
            }
            else
            {
                tail->next=cur1;
                tail=tail->next;
            }
            cur1=cur1->next;
        }
        else
        {
             if (head==NULL)
            {
                head=tail=cur2;
            }
            else
            {
                tail->next=cur2;
                tail=tail->next;
            }
            cur2=cur2->next;           
        }
    }
    
    if(cur1)
        tail->next=cur1;
    if(cur2)
        tail->next=cur2;
        
    return head;
}

分析

尾插要有尾指针tail(这样不用频繁找尾),同时要有指向头节点的指针head用于返回

cur1->val<cur2->val和cur1->val>=cur2->val操作方式是类似的