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 = None

class Solution:
    # @param {ListNode} l1
    # @param {ListNode} l2
    # @return {ListNode}
    def mergeTwoLists(self, l1, l2):
        if l1==None:return l2
        if l2==None:return l1
        p=l1
        if l1.val > l2.val:
            p=l2;l2=l2.next
        else:p=l1;l1=l1.next
        head=p
        while l1!=None and l2!=None:
            if l1.val > l2.val:
                p.next=l2;l2=l2.next
            else:p.next=l1;l1=l1.next
            p=p.next
        if l1==None:p.next=l2
        else:p.next=l1
        return head