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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：

输入：l1 = [1,2,4], l2 = [1,3,4]
输出：[1,1,2,3,4,4]

示例 2：

输入：l1 = [], l2 = []
输出：[]

示例 3：

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

• /*** 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:

/*** 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

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

完整代码

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)

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