Leetcode刷題-merged two sorted list

題目：merged two sorted list

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.
Example:
Input: 1->2->4, 1->3->4
Output: 1->1->2->3->4->4

---

解法：
python3 直接迭代

Definition for singly-linked list.
class ListNode:
def init(self, x):
self.val = x
self.next = None
class Solution:
def mergeTwoLists(self, l1, l2):
“”"
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
“”"
if not l1 or not l2:
return l1 or l2
re=cur = ListNode(0)
while(l1 and l2):
if l1.val<l2.val:
cur.next=ListNode(l1.val)
cur=cur.next
l1=l1.next
else:
cur.next=ListNode(l2.val)
cur=cur.next
l2=l2.next
if not l1 or not l2:
cur.next=l1 or l2
return re.next

---

c++ 直接迭代

• 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 and l2 NULL )
  return NULL ;
  else if (l1NULL)
  {
  return l2;
  }
  else if (l2NULL)
  {
  return l1;
  }
  else
  {
  ListNode cur{-1};
  ListNode* re =&cur ;
  ListNode* res=re;
  while (l1 and l2 )
  {
  // l1=l1->next;
  // l2=l2->next;
  if (l1->val < l2->val)
  {
  re->next = l1;
  l1=l1->next;
  re=re->next;
  }
  else{
  re->next=l2;
  l2=l2->next;
  re=re->next;
  }
  }
  if (l1==NULL)
  re->next=l2;
  else{
  re->next=l1;
  }
  return res->next;
  }
  }
  };