合并两个排序的链表(递归解法需复习)

image.png

解法一、非递归
看到这个问题，想到了最熟悉的归并排序中的归并的过程，其实两个过程是一样的。可以完全类比。

1. 首先判断是否为空链表，如果一方为空链表，则直接返回另外的头指针即可。
2. 然后合并两个链表，将两个链表的头部小的那个，作为p指针的next, 直到有一方空了，直接将另一方接到后面就好了。

# -*- 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:
            head = pHead1
            pHead1 = pHead1.next
        else:
            head = pHead2
            pHead2 = pHead2.next
        p = head
        while pHead1 and pHead2:
            if pHead1.val <= pHead2.val:
                p.next = pHead1
                pHead1 = pHead1.next
                p = p.next
            else:
                p.next = pHead2
                pHead2 = pHead2.next
                p = p.next
        if pHead1 == None:
            p.next = pHead2
        if pHead2 == None:
            p.next = pHead1
        return head

解法二：递归的方法

image.png

# -*- 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
        elif pHead2 == None:
            return pHead1
        
        pMergedHead = None
        if pHead1.val < pHead2.val:
            pMergedHead = pHead1
            pMergedHead.next = self.Merge(pHead1.next,pHead2)
        else:
            pMergedHead = pHead2
            pMergedHead.next = self.Merge(pHead1, pHead2.next)
        return pMergedHead