合并两个排序的链表的解题思路
合并两个排序的链表
https://www.nowcoder.com/practice/d8b6b4358f774294a89de2a6ac4d9337?tpId=295&tags=&title=&difficulty=0&judgeStatus=0&rp=0&sourceUrl=%2Fexam%2Foj%3FquestionJobId%3D10%26subTabName%3Donline_coding_page%26topicNameBigMember%3D%25E9%259D%25A2%25E8%25AF%2595TOP101%26topicIdBigMember%3D295%26pageSourceBigMember%3Dnc03%26dayCountBigMember%3D%25E8%25BF%259E%25E7%25BB%25AD%25E5%258C%2585%25E6%259C%2588
思路如下：
- 定义3个指针变量，用于指向链表一和链表二，一个用于指向定义的头节点；
- 定义一个头节点，用于指向比较完成后的第一个节点；
- 将链表一和链表二的指针进行比较，头节点的指针指向小的那一个节点，链表指针后移，头节点指针后移
- 将剩下的链表直接连接
- 返回头节点的下一个节点，目的去掉头节点
struct ListNode* Merge(struct ListNode* pHead1, struct ListNode* pHead2 ) {
if(pHead1==NULL){
return pHead2;
}
if(pHead2==NULL){
return pHead1;
}
struct ListNode dummy;
struct ListNode *tail = &dummy;
while(pHead1!=NULL&&pHead2!=NULL){
if(pHead1->val<pHead2->val){
tail->next = pHead1;
pHead1 = pHead1->next;
}
else{
tail->next = pHead2;
pHead2 = pHead2->next;
}
tail = tail->next;
}
if(pHead1!=NULL){
tail->next = pHead1;
}
if(pHead2!=NULL){
tail->next = pHead2;
}
return tail->next;
}
