[Leetcode21]合并两个有序链表

将两个有序链表合并为一个新的有序链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。

这题比较简单，一定要记住备份表头用来返回链表。

代码如下，还可以精简：

python：

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def mergeTwoLists(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        begin = ListNode(0)
        res= begin
        if l1 == None and l2==None:
            return None
        elif l1 != None and l2==None:
            return l1
        elif l1 == None and l2!=None:
            return l2
        while l1 or l2:
            if l1.val <= l2.val:
                temp = ListNode(0)
                res.next = temp
                temp.val = l1.val
                res = temp
                if l1.next:
                    l1 = l1.next
                else:
                    res.next = l2
                    return begin.next
            else:
                temp = ListNode(0)
                res.next = temp
                temp.val = l2.val
                res = temp
                if l2.next:
                    l2 = l2.next
                else:
                    res.next = l1
                    return begin.next

C++: 

/**
 * 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) {
        if(l1 == NULL && l2 == NULL) return NULL;
        else if(l1 != NULL && l2 == NULL) return l1;
        else if(l1 == NULL && l2 != NULL) return l2;
        ListNode* begin = new ListNode(0);
        ListNode* res = begin;
        while(l1!=NULL || l2!=NULL){
            if(l1->val <= l2->val){
                ListNode* temp = new ListNode(0);
                res->next = temp;
                temp->val = l1->val;
                res = temp;
                if(l1->next != NULL) l1 = l1->next;
                else{
                    temp->next = l2;
                    return begin->next;
                }
            }
            else{
                ListNode* temp = new ListNode(0);
                res->next = temp;
                temp->val = l2->val;
                res = temp;
                if(l2->next != NULL) l2 = l2->next;
                else{
                    temp->next = l1;
                    return begin->next;
                }
            }
        }
    }
};