1. 题目信息

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。

示例：输入：1->2->4, 1->3->4
输出：1->1->2->3->4->4

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/merge-two-sorted-lists
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2. 解题

• 链表的基本操作，注意把前一个节点和下一个节点（l1,l2当前位置里的较小者）连起来

/*** 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)return l2;else if(l2 == NULL)return l1;else{ListNode *newHead = new ListNode(0);ListNode *curNode = newHead->next, *prev = newHead;while(l1 && l2){if(l1->val < l2->val){curNode = l1;l1 = l1->next;}else{curNode = l2;l2 = l2->next;}prev->next = curNode;prev = curNode;curNode = curNode->next;}if(l1 == NULL){prev->next = l2;}else//l2 == NULL{prev->next = l1;}return newHead->next;}        }
};

class Solution {	//2020.5.1
public:ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {if(!l1) return l2;if(!l2) return l1;ListNode* empty = new ListNode(0);ListNode* cur = empty;while(l1 && l2){if(l1->val <= l2->val){cur->next = l1;l1 = l1->next;}else{cur->next = l2;l2 = l2->next;}cur = cur->next;}cur->next = (l1 ? l1 : l2);return empty->next;}
};

《剑指Offer》同题：面试题25. 合并两个排序的链表

class Solution {
public:ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {if(!l1)return l2;if(!l2)return l1;ListNode* emptyHead = new ListNode(0);ListNode* cur = emptyHead;while(l1 && l2){while(l1 && l2 && l1->val <= l2->val){cur->next = l1;cur = l1;l1 = l1->next;}while(l1 && l2 && l1->val > l2->val){cur->next = l2;cur = l2;l2 = l2->next;}}if(!l1)cur->next = l2;elsecur->next = l1;return emptyHead->next;}
};