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。

思路

《剑指offer》P114
新建一个节点pre作为头结点，取list1和list2中当前值较小的结点接入，并将指针后移，直到其中一个链表到达末尾。

code

• 非递归

# -*- 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 not pHead1:
            return pHead2
        if not pHead2:
            return pHead1
        pre = ListNode(-1)
        pCurr = pre
        while pHead1 and pHead2:
            if pHead1.val <= pHead2.val:
                pCurr.next = pHead1
                pHead1 = pHead1.next
            else:
                pCurr.next = pHead2
                pHead2 = pHead2.next
            pCurr = pCurr.next
        if pHead1:
            pCurr.next = pHead1
        if pHead2:
            pCurr.next = pHead2
        return pre.next

• 递归

# -*- 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 not pHead1:
            return pHead2
        if not pHead2:
            return pHead1
        mergeHead = None
        if (pHead1.val <= pHead2.val):
            mergeHead = pHead1
            mergeHead.next = self.Merge(pHead1.next, pHead2)
        else:
            mergeHead = pHead2
            mergeHead.next = self.Merge(pHead1, pHead2.next)
        return mergeHead