[剑指Offer] 16.合并两个排序链表

题目描述

输入两个单调递增的链表，输出两个链表合成后的链表，当然我们需要合成后的链表满足单调不减规则。

【思路1】递归

 /*
 struct ListNode {
     int val;
     struct ListNode *next;
     ListNode(int x) :
             val(x), next(NULL) {
     }
 };*/
 class Solution {
 public:
     ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
     {
         if(pHead1 == NULL)
             return pHead2;
         else if(pHead2 == NULL)
             return pHead1;
         ListNode* res = NULL;
         if(pHead1->val <= pHead2->val){
             res = pHead1;
             res->next = Merge(pHead1->next, pHead2);
         }else{
             res = pHead2;
             res->next = Merge(pHead1,pHead2->next);
         }
         return res;
     }
 };

【思路2】非递归，新建一个链表并保存头结点，将原来两个链表进行比较按顺序插入到新链表中，最后将有剩余的链表直接接上。

 /*
 struct ListNode {
     int val;
     struct ListNode *next;
     ListNode(int x) :
             val(x), next(NULL) {
     }
 };*/
 class Solution {
 public:
     ListNode* Merge(ListNode* pHead1, ListNode* pHead2){
         if(pHead1 == NULL)
             return pHead2;
         else if(pHead2 == NULL)
             return pHead1;
         ListNode* res = NULL;
         ListNode* cur = NULL;
         while(pHead1 != NULL && pHead2 != NULL){
             if(pHead1->val <= pHead2->val){
                 if(res == NULL)
                     res = cur = pHead1;
                 else{
                     cur->next = pHead1;
                     cur = cur->next;
                 }
                 pHead1 = pHead1->next;
             }else{
                 if(res == NULL)
                     res = cur = pHead2;
                 else{
                     cur->next = pHead2;
                     cur = cur->next;
                 }
                 pHead2 = pHead2->next;
             }
         }
         if(pHead1 == NULL){
             cur->next = pHead2;
         }
         if(pHead2 == NULL){
             cur->next = pHead1;
         }
         return res;
     }
 };