LeetCode OJ：Merge Two Sorted Lists（合并两个链表）
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.
注意题目要求合并的时候不能新建节点，直接使用原来的节点，比较简单，代码如下：
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution { 10 public: 11 ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) { 12 if(l1 == NULL) return l2; 13 if(l2 == NULL) return l1; 14 ListNode * root = new ListNode(-1); 15 ListNode * helper = root; 16 while(l1!=NULL && l2!=NULL){ 17 if(l1->val <= l2->val) 18 root->next = l1, l1=l1->next; 19 else if(l1->val > l2->val) 20 root->next = l2, l2=l2->next; 21 root = root->next; 22 } 23 while(l1!=NULL){ 24 root->next = l1; 25 l1 = l1->next; 26 root = root->next; 27 } 28 while(l2!=NULL){ 29 root->next = l2; 30 l2 = l2->next; 31 root = root->next; 32 } 33 return helper->next; 34 } 35 };
1 public class Solution { 2 public ListNode mergeTwoLists(ListNode l1, ListNode l2) { 3 ListNode helper = new ListNode(0); 4 ListNode ret = helper; 5 if(l1 == null) return l2; 6 if(l2 == null) return l1; 7 while(l1 != null && l2 != null){ 8 if(l1.val < l2.val){ 9 helper.next = l1; 10 l1 = l1.next; 11 }else{ 12 helper.next = l2; 13 l2 = l2.next; 14 } 15 helper = helper.next; 16 } 17 while(l1 != null){ 18 helper.next = l1; 19 l1 = l1.next; 20 helper = helper.next; 21 } 22 while(l2 != null){ 23 helper.next = l2; 24 l2 = l2.next; 25 helper = helper.next; 26 } 27 return ret.next; 28 } 29 }
