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
依次遍历两个链表，比较两个链表的元素，采用尾插法，小的先插入链表，大的后插入链表
代码# -*- 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#注意将待插入链表头后移
if (pHead1 != None):
tmp.next = pHead1
if (pHead2 != None):
tmp.next = pHead2
return phNew.next
注意
错误代码：# -*- 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:#若其中一个为空，另一个不为空，也是可以满足题意的。见代码下方的测试用例
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
测试用例:
{1,3,5},{}
对应输出应该为:
{1,3,5}
你的输出为:
{}
)
