[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

 1 /* 2 struct ListNode { 3     int val; 4     struct ListNode *next; 5     ListNode(int x) : 6             val(x), next(NULL) { 7     } 8 };*/ 9 class Solution {10 public:11     ListNode* Merge(ListNode* pHead1, ListNode* pHead2)12     {13         if(pHead1 == NULL)14             return pHead2;15         else if(pHead2 == NULL)16             return pHead1;17         ListNode* res = NULL;18         if(pHead1->val <= pHead2->val){19             res = pHead1;20             res->next = Merge(pHead1->next, pHead2);21         }else{22             res = pHead2;23             res->next = Merge(pHead1,pHead2->next);24         }25         return res;26     }27 };

 【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。

 1 /* 2 struct ListNode { 3     int val; 4     struct ListNode *next; 5     ListNode(int x) : 6             val(x), next(NULL) { 7     } 8 };*/ 9 class Solution {10 public:11     ListNode* Merge(ListNode* pHead1, ListNode* pHead2){12         if(pHead1 == NULL)13             return pHead2;14         else if(pHead2 == NULL)15             return pHead1;16         ListNode* res = NULL;17         ListNode* cur = NULL;18         while(pHead1 != NULL && pHead2 != NULL){19             if(pHead1->val <= pHead2->val){20                 if(res == NULL)21                     res = cur = pHead1;22                 else{23                     cur->next = pHead1;24                     cur = cur->next;25                 }26                 pHead1 = pHead1->next;27             }else{28                 if(res == NULL)29                     res = cur = pHead2;30                 else{31                     cur->next = pHead2;32                     cur = cur->next;33                 }34                 pHead2 = pHead2->next;35             }36         }37         if(pHead1 == NULL){38             cur->next = pHead2;39         }40         if(pHead2 == NULL){41             cur->next = pHead1;42         }43         return res;44     }45 };