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.
1为了操作方便，我们给拼接后的新链表添加一个头结点 本文地址
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/**
* 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) {
//为操作方便，添加一个头节点
ListNode node(0), *p = &node;
while
(l1 && l2)
{
if
(l1->val < l2->val)
{
p->next = l1;
l1 = l1->next;
}
else
{
p->next = l2;
l2 = l2->next;
}
p = p->next;
}
if
(l1)p->next = l1;
else
if
(l2)p->next = l2;
return
node.next;
}
};
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本文转自tenos博客园博客，原文链接：http://www.cnblogs.com/TenosDoIt/p/3673050.html，如需转载请自行联系原作者
