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Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.
Example:
Input: 1->2->4, 1->3->4 Output: 1->1->2->3->4->4
合并两个有序链表，最后返回新的链表。实现思路很简单，对比两个链表每一个结点值的大小，然后优先拼接到新表的尾部。
1.非递归方法:
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * struct ListNode *next; 6 * }; 7 */ 8 struct ListNode* mergeTwoLists(struct ListNode* l1, struct ListNode* l2) { 9 if(l1 == NULL) return l2; 10 if(l2 == NULL) return l1; 11 if(l1==NULL && l2==NULL) return NULL; 12 struct ListNode *head=NULL; 13 if(l1->val < l2->val) 14 { 15 head=l1; 16 l1=l1->next; 17 } 18 else 19 { 20 head=l2; 21 l2=l2->next; 22 } 23 struct ListNode *p=head; 24 while(l1 && l2) 25 { 26 if(l1->val < l2->val) 27 { 28 p->next=l1; 29 l1=l1->next; 30 } 31 else 32 { 33 p->next=l2; 34 l2=l2->next; 35 } 36 p=p->next; 37 } 38 39 if(l1) p->next=l1; //如果还剩下l1结点，则把l1加到新链表 40 if(l2) p->next=l2; //如果还剩下l2结点，则把l2加到新链表 41 return head; 42 }
2.递归方法。这个方法还是看别的大神做的，代码很简洁，很惊艳。
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * struct ListNode *next; 6 * }; 7 */ 8 struct ListNode* mergeTwoLists(struct ListNode* l1, struct ListNode* l2) { 9 if(l1 == NULL) return l2; 10 if(l2 == NULL) return l1; 11 if(l1==NULL && l2==NULL) return NULL; 12 //假如一开始是l1数据域小，递归结束后返回新l1(即首地址),反之同理。 13 if(l1->val <= l2->val){ 14 l1->next = mergeTwoLists(l1->next,l2); 15 return l1; 16 }else{ 17 l2->next = mergeTwoLists(l1,l2->next); 18 return l2; 19 } 20 }