剑指offer16-合并两个排序的链表

题目描述

输入两个单调递增的链表，输出两个链表合成后的链表，当然我们需要合成后的链表满足单调不减规则。

示例

输入      {1,3,5},{2,4,6}

返回值  {1,2,3,4,5,6}

知识点回顾

排序链表合并

代码

#递归法
# -*- coding:utf-8 -*-
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
class Solution:
    # 返回合并后列表
    def Merge(self, pHead1, pHead2):
        # write code here
        if pHead1==None:
            return pHead2
        if pHead2==None:
            return pHead1
        if pHead1.val<pHead2.val:
            newHead=pHead1
            pHead1=pHead1.next
            newHead.next=self.Merge(pHead1, pHead2)
        else:
            newHead=pHead2
            pHead2=pHead2.next
            newHead.next=self.Merge(pHead1, pHead2)
        return newHead

#递归比较复杂，可以搭配debug：
a1=ListNode(1)
b1=ListNode(3)
c1=ListNode(5)
a2=ListNode(2)
b2=ListNode(4)
c2=ListNode(6)
a1.next=b1
b1.next=c1
a2.next=b2
b2.next=c2
c=Solution()
c.Merge(a1,a2)

#非递归法
# -*- coding:utf-8 -*-
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
class Solution:
    # 返回合并后列表
    def Merge(self, pHead1, pHead2):
        newHead=ListNode(0)                     #随便找个头节点
        tmp=newHead                             #记录当前插入位置的指针
        while pHead1 and pHead2:
            if pHead1.val<pHead2.val:
                tmp.next=pHead1
                pHead1=pHead1.next
            else:
                tmp.next=pHead2
                pHead2=pHead2.next
            tmp=tmp.next
        if pHead1:
            tmp.next=pHead1
        if pHead2:
            tmp.next=pHead2
        return newHead.next