Leetcode 21.合并两个有序链表

21.合并两个有序链表

题目

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

方法

方法1

# 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
        """
        ptr1 = l1
        ptr2 = l2
        head = ListNode(0)
        tail =head
        while ptr1 and ptr2:
            if ptr1.val < ptr2.val:
                tail.next = ptr1
                tail = ptr1
                ptr1 = ptr1.next  # 这两句不换调换顺序
                tail.next = None  # 这两句不换调换顺序
            else:
                tail.next = ptr2
                tail =ptr2
                ptr2 = ptr2.next  # 这两句不换调换顺序
                tail.next = None  # 这两句不换调换顺序
        if ptr1:
            tail.next = ptr1
        elif ptr2:
            tail.next = ptr2
        return head.next

# 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
        """
        head=ListNode(0)
        tail=head
        while l1 and l2:
            if l1.val<l2.val:
                tail.next=l1
                tail=l1

                l1=l1.next
                tail.next=None
            else:
                tail.next=l2
                tail=l2

                l2=l2.next
                tail.next=None
        if l1:
            tail.next=l1
        elif l2:
            tail.next=l2
        return head.next

方法2：递归

# 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
        elif not l2:
            return l1

        mergeHead = ListNode(0)

        if l1.val < l2.val:
            mergeHead = l1
            mergeHead.next = self.mergeTwoLists(l1.next, l2)
        else:
            mergeHead = l2
            mergeHead.next = self.mergeTwoLists(l1,l2.next)

        return mergeHead