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.
Example:
Input: 1->2->4, 1->3->4
Output: 1->1->2->3->4->4
思路:
两链表不一定等长,先遍历比较两链表都不为空时的最小值存入新链表,当某一链表下一个值为空时,将另一链表剩余部分存入新链表
当前解决:
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def mergeTwoLists(self, l1, l2):
l3 = ListNode(0)
p = l3
while l1 and l2:
if l1.val < l2.val:
p.next = l1
l1 = l1.next
else:
p.next = l2
l2 = l2.next
p = p.next
if l1:
p.next = l1
else:
p.next = l2
return l3.next