原题
给你一个链表以及两个权值v1和v2，交换链表中权值为v1和v2的这两个节点。保证链表中节点权值各不相同，如果没有找到对应节点，那么什么也不用做。
解题思路
- 基础的链表操作，写一个helper函数，根据head和value找出和value相等的节点和prev节点
- 比如给出3->4->1->5->Null 和 1 返回Node(4)和Node(1)
- 注意边界情况，如果v1=4, v2=1 那么current1和pre2重合，要特殊考虑
完整代码
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
# @param {ListNode} head, a ListNode
# @oaram {int} v1 an integer
# @param {int} v2 an integer
# @return {ListNode} a new head of singly-linked list
def swapNodes(self, head, v1, v2):
# Write your code here
dummy = ListNode(0)
dummy.next = head
prev1, cur1 = self.findNode(dummy, v1)
prev2, cur2 = self.findNode(dummy, v2)
if cur1 and cur2:
prev1.next = cur2
temp = cur2.next
if cur1.next == cur2:
cur2.next = cur1
cur1.next = temp
else:
cur2.next = cur1.next
prev2.next = cur1
cur1.next = temp
return dummy.next
def findNode(self, head, value):
while head.next != None:
if head.next.val == value:
return head, head.next
head = head.next
return None, None
