题目：
将两个升序链表合并为一个新的 升序 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
思路：
注意输入空链表时的特殊处理；
使用两个指针分别指向两个链表；
合并过程中，每次合并的步骤都和之前的相同，因此可以采用递归解法，递归解法的思路需要注意。
Python解法：
1 class ListNode(object): 2 def __init__(self, x): 3 self.val = x 4 self.next = None 5 6 7 class Solution(object): 8 def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode: 9 if l1 is None: 10 return l2 11 elif l2 is None: 12 return l1 13 prehead = ListNode(0) # 定义一个不存储任何数据的哨兵节点,便于返回结果 14 prev = prehead 15 while l1 and l2: 16 if l1.val < l2.val: 17 prev.next = l1 18 l1 = l1.next 19 else: 20 prev.next = l2 21 l2 = l2.next 22 prev = prev.next 23 if l1 is None: 24 prev.next = l2 25 if l2 is None: 26 prev.next = l1 27 return prehead.next 28 29 30 if __name__ == '__main__': 31 node_11 = ListNode(1) 32 node_12 = ListNode(3) 33 node_13 = ListNode(4) 34 node_11.next = node_12 35 node_12.next = node_13 36 37 node_21 = ListNode(0) 38 node_22 = ListNode(2) 39 node_23 = ListNode(5) 40 node_21.next = node_22 41 node_22.next = node_23 42 43 s = Solution() 44 newNode = s.mergeTwoLists(node_11, node_21) 45 while newNode.next is not None: 46 print(newNode.val) 47 newNode = newNode.next
C++递归解法：
1 struct ListNode { 2 int val; 3 ListNode *next; 4 ListNode(int x) : val(x), next(NULL) {} 5 }; 6 7 class Solution 8 { 9 public: 10 ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) { 11 if (l1 == NULL) 12 return l2; 13 if (l2 == NULL) 14 return l1; 15 ListNode *tempHead = new ListNode(-1); 16 17 if (l1->val < l2->val) { 18 tempHead = l1; 19 tempHead->next = mergeTwoLists(l1->next, l2); 20 } 21 else { 22 tempHead = l2; 23 tempHead->next = mergeTwoLists(l1, l2->next); 24 } 25 return tempHead; // tempHead会一级一级的返回 26 } 27 }; 28 29 int main() { 30 ListNode *node_11 = new ListNode(1); 31 ListNode *node_12 = new ListNode(3); 32 ListNode *node_13 = new ListNode(5); 33 node_11 -> next = node_12; 34 node_12->next = node_13; 35 36 ListNode *node_21 = new ListNode(2); 37 ListNode *node_22 = new ListNode(4); 38 ListNode *node_23 = new ListNode(4); 39 node_21->next = node_22; 40 node_22->next = node_23; 41 42 Solution s; 43 ListNode *newNode = s.mergeTwoLists(node_11, node_21); 44 while (newNode != NULL) { 45 cout << newNode->val << " "; 46 newNode = newNode->next; 47 }