一. 题目描述

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.

二. 题目分析

合并两个有序链表，递归或迭代均可以解决。

三. 示例代码

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

四. 小结

该题需要一些边界条件判断。