Problem:
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.
Solution:
两个有序链表，每次取头部最小的那个元素，然后将这个元素从原来链表中去掉，依次循环
题目大意：
给定两个有序链表，要求将链表合并为一个有序链表，并且不能使用额外的空间
Java源代码（344ms）：
/*** Definition for singly-linked list.* public class ListNode {* int val;* ListNode next;* ListNode(int x) { val = x; }* }*/
public class Solution {public ListNode mergeTwoLists(ListNode l1, ListNode l2) {ListNode head,p;if(l1==null)return l2;if(l2==null)return l1;if(l1.val > l2.val){p=l2;l2=l2.next;}else{p=l1;l1=l1.next;}head=p;while(l1!=null && l2!=null){if(l1.val > l2.val){p.next=l2;l2=l2.next;}else{p.next=l1;l1=l1.next;}p=p.next;}if(l1==null)p.next=l2;else p.next=l1;return head;}
}C语言源代码（3ms）：
/*** Definition for singly-linked list.* struct ListNode {* int val;* struct ListNode *next;* };*/
struct ListNode* mergeTwoLists(struct ListNode* l1, struct ListNode* l2) {struct ListNode *head,*p;if(l1==NULL)return l2;if(l2==NULL)return l1;if(l1->val > l2->val){p=l2;l2=l2->next;}else{p=l1;l1=l1->next;}head=p;while(l1!=NULL && l2!=NULL){if(l1->val > l2->val){p->next=l2;l2=l2->next;}else{p->next=l1;l1=l1->next;}p=p->next;}if(l1==NULL)p->next=l2;else p->next=l1;return head;
}C++源代码（13ms）：
/*** Definition for singly-linked list.* struct ListNode {* int val;* ListNode *next;* ListNode(int x) : val(x), next(NULL) {}* };*/
class Solution {
public:ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {ListNode *head,*p;if(l1==NULL)return l2;if(l2==NULL)return l1;if(l1->val > l2->val){p=l2;l2=l2->next;}else{p=l1;l1=l1->next;}head=p;while(l1!=NULL && l2!=NULL){if(l1->val > l2->val){p->next=l2;l2=l2->next;}else{p->next=l1;l1=l1->next;}p=p->next;}if(l1==NULL)p->next=l2;else p->next=l1;return head;}
};Python源代码（67ms）：
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = Noneclass Solution:# @param {ListNode} l1# @param {ListNode} l2# @return {ListNode}def mergeTwoLists(self, l1, l2):if l1==None:return l2if l2==None:return l1p=l1if l1.val > l2.val:p=l2;l2=l2.nextelse:p=l1;l1=l1.nexthead=pwhile l1!=None and l2!=None:if l1.val > l2.val:p.next=l2;l2=l2.nextelse:p.next=l1;l1=l1.nextp=p.nextif l1==None:p.next=l2else:p.next=l1return head