1 问题
链表实现选择排列中经常会遇到一些问题，那么该如何解决它们呢？
2 方法
这一类问题的基本都是根据题目给定的条件，对链表进行各种组合，如：基于归并排序思想，根据节点的数值，合并两个链表（合并两个排序的链表、合并k个已排序的链表）根据节点的位置，对链表重新排序（链表的奇偶重排）对两个链表节点的数值相加（链表相加(二))
假设链表中每一个节点的值都在 0 - 9 之间，那么链表整体就可以代表一个整数。给定两个这种链表，请生成代表两个整数相加值的结果链表。 整体思路，如题目，链表的顺序与加法的顺序是相反的，自然的想到两种思路：把链表的元素压入栈中，借助栈实现对反转链表的元素进行操作；直接反转链表由于两种方式都需要新建链表，存储两个整数的相加值，因此空间复杂度都是o(n)。方法1比2多一个栈的空间，但是总的空间复杂度也是o(n)。细节提示加法的10进制的进位。设置进位标志incre，每次循环判断 val1 = list1.pop(-1) list2.pop(-1) incre。并且，在循环结束后，需要判断incre是否>0，如果>0，需要在链表中增加
代码清单 1
代码语言：javascript复制class ListNode:
def __init__(self, x):
self.val = x
self.next = None
class Solution:
def addInList(self , head1 , head2 ):
# write code here
list1 = []
while head1:
list1.append(head1.val)
head1 = head1.next
list2 = []
while head2:
list2.append(head2.val)
head2 = head2.next
list3 = []
incre = 0
while len(list1) and len(list2):
val1 = list1.pop(-1) list2.pop(-1) incre
incre = val1/10
val1 = val1
list3.append(val1)
while len(list1):
val1 = list1.pop(-1) incre
incre = val1/10
val1 = val1
list3.append(val1)
while len(list2):
val1 = list2.pop(-1) incre
incre = val1/10
val1 = val1
list3.append(val1)
if incre>0:
list3.append(incre)
dumpyNode = ListNode(-1)
pHead = dumpyNode
while len(list3):
pHead.next = ListNode(list3.pop(-1))
pHead = pHead.next
return dumpyNode.next
def addInList2(self , head1 , head2 ):
cur1 = head1
pre = None
while cur1:
next1 = cur1.next
cur1.next = pre
pre = cur1
cur1 = next1
head1 = pre
cur2 = head2
pre2 = None
while cur2:
next2 = cur2.next
cur2.next = pre2
pre2 = cur2
cur2 = next2
head2 = pre2
dumpyNode3 = ListNode(-1)
pHead = dumpyNode3
incre = 0
while head1 and head2:
val = head1.val head2.val incre
incre = val/10
val = val
head = ListNode(val)
pHead.next = head
pHead = pHead.next
head1 = head1.next
head2 = head2.next
while head1:
val = head1.val incre
incre = val/10
val = val
head = ListNode(val)
pHead.next = head
pHead = pHead.next
head1 = head1.next
while head2:
val = head2.val incre
incre = val/10
val = val
head = ListNode(val)
pHead.next = head
pHead = pHead.next
head2 = head2.next
if incre>0:
head = ListNode(incre)
pHead.next = head
pHead = pHead.next
pHead = dumpyNode3.next
cur1 = pHead
pre = None
while cur1:
next1 = cur1.next
cur1.next = pre
pre = cur1
cur1 = next1
return pre3 结语
针对数组排序问题，提出的解决方法，证明该方法是有效的。其实上面的题目的思路都很简单，相当于把简单的排序从数组迁移到了链表中。个人认为技巧在于链表节点的生成与穿针引线，一般可以使用两个辅助节点，定义虚拟节点和游走节点，虚拟节点负责返回整个链表，游走节点负责穿针引线。以提高算法效率。
