題目描述

輸入兩個(gè)單調(diào)遞增的鏈表，輸出兩個(gè)鏈表合成后的鏈表，當(dāng)然我們需要合成后的鏈表滿足單調(diào)不減規(guī)則。

思路

依次遍歷兩個(gè)鏈表，比較兩個(gè)鏈表的元素，采用尾插法，小的先插入鏈表，大的后插入鏈表

代碼# -*- 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

phNew = ListNode(0);

tmp = phNew

while pHead1 != None and pHead2 != None:

if pHead1.val > pHead2.val:

tmp.next = pHead2

pHead2 = pHead2.next

else:

tmp.next = pHead1

pHead1 = pHead1.next

tmp = tmp.next#注意將待插入鏈表頭后移

if (pHead1 != None):

tmp.next = pHead1

if (pHead2 != None):

tmp.next = pHead2

return phNew.next

注意

錯(cuò)誤代碼：# -*- 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 or pHead2 == None:#若其中一個(gè)為空，另一個(gè)不為空，也是可以滿足題意的。見(jiàn)代碼下方的測(cè)試用例

return

phNew = ListNode(0);

tmp = phNew

while pHead1 != None and pHead2 != None:

if pHead1.val > pHead2.val:

tmp.next = pHead2

pHead2 = pHead2.next

else:

tmp.next = pHead1

pHead1 = pHead1.next

tmp = tmp.next

if (pHead1 != None):

tmp.next = pHead1

if (pHead2 != None):

tmp.next = pHead2

return phNew.next

測(cè)試用例:

{1,3,5},{}

對(duì)應(yīng)輸出應(yīng)該為:

{1,3,5}

你的輸出為:

{}

總結(jié)

以上是生活随笔為你收集整理的链表 python 牛客_牛客网《剑指offer》之Python2.7实现：合并两个排序的链表的全部?jī)?nèi)容，希望文章能夠幫你解決所遇到的問(wèn)題。

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