LeetCode 21

一、问题描述

Description:

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.

合并两个有序的链表，返回合并后的链表。

二、解题报告

本题也很简单，按顺序不断取下 两个链表表头中 较小的结点放入新的链表中。然后看哪个链表还有剩余，将剩下的部分加到新链表的后面。

代码如下：

class Solution {
public:
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        ListNode* head = new ListNode(0);
        ListNode* p = head;
        while(l1!=NULL && l2!=NULL) {
            if(l1->val < l2->val) {
                p->next = l1;
                l1 = l1->next;
            }
            else {
                p->next = l2;
                l2 = l2->next;
            }
            p = p->next;
            p->next = NULL;
        }

        if(l1!=NULL)
            p->next = l1;
        if(l2!=NULL)
            p->next = l2;

        return head->next;
    }
};

LeetCode答案源代码：https://github.com/SongLee24/LeetCode