21. Merge Two Sorted Lists [Easy] [Linked List]
题意
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.
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。
Example:
Input: 1->2->4, 1->3->4 Output: 1->1->2->3->4->4
思路一
合并后的链表仍然是有序的，可以同时遍历两个链表，每次选取两个链表中较小值的节点，依次连接起来，就能得到最终的链表。
# 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): """ :type l1: ListNode :type l2: ListNode :rtype: ListNode """ if not l1 or not l2: return l1 or l2 head = cur = ListNode(0) while l1 and 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 or l2 return head.next
思路二
递归的思想。
# 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): """ :type l1: ListNode :type l2: ListNode :rtype: ListNode """ if not l1: return l2 if not l2: return l1 if l1.val < l2.val: l1.next = self.mergeTwoLists(l1.next,l2) return l1 else: l2.next = self.mergeTwoLists(l1,l2.next) return l2