题目要求: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.
简单的说就是将量和排好序的链表合并为一个链表。这道题的考察点为链表的操作。需要注意的是对链表有效性的判断,代码主要分为三部分,第一部分对输入指针的判断,第二部分是根据链表节点大小进行合并,第三部分是当一个链表达到最后节点之后,将没达到最后节点的链表链接到返回链表的后面,这样可以减少后面部分的循环操作。需要注意的是对于返回链表需要保存其头指针。
/**
* 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 l2;
else if (l1 != NULL && l2 == NULL)
return l1;
ListNode* l = NULL;
ListNode* head = NULL;
if (l1->val < l2->val)
{
l = l1;
l1 = l1->next;
}
else
{
l = l2;
l2 = l2->next;
}
head = l;
while (l1 != NULL && l2 != NULL)
{
if (l1->val < l2->val)
{
l->next = l1;
l1 = l1->next;
l = l->next;
}
else
{
l->next = l2;
l2 = l2->next;
l = l->next;
}
}
if (l1 != NULL)
l->next = l1;
else if (l2 != NULL)
l->next = l2;
return head;
}
};
Leetcode (2) Merge Two Sorted Lists
原文地址:http://blog.csdn.net/angelazy/article/details/44995369