[Leetcode][Python]24: Swap Nodes in Pairs
# -*- coding: utf8 -*-'''
__author__ = 'dabay.wang@gmail.com'
24: Swap Nodes in Pairs
https://oj.leetcode.com/problems/swap-nodes-in-pairs/
Given a linked list, swap every two adjacent nodes and return its head.
For example,
Given 1->2->3->4, you should return the list as 2->1->4->3.
Your algorithm should use only constant space. You may not modify the values in the list, only nodes itself can be changed.
===Comments by Dabay===
链表的操作。主要就是赋值next的时候，如果这个next本来指向的node还有用，就把它先记录下来。
这里用previous指向要交换的一对node的前面一个node。要交换的node依次为node1和node2。
'''
# Definition for singly-linked list.
class ListNode:
def __init__(self, x):
self.val = x
self.next = None
class Solution:
# @param a ListNode
# @return a ListNode
def swapPairs(self, head):
previous = new_head = ListNode(0)
new_head.next = head
while previous and previous.next and previous.next.next:
node1 = previous.next
node2 = node1.next
node1.next = node2.next
previous.next = node2
node2.next = node1
previous = node1
return new_head.next
def print_listnode(node):
while node:
print "%s->" % node.val,
node = node.next
print "END"
def main():
sol = Solution()
node = root = ListNode(1)
for i in xrange(2, 5):
node.next = ListNode(i)
node = node.next
print_listnode(root)
print_listnode(sol.swapPairs(root))
if __name__ == '__main__':
import time
start = time.clock()
main()
print "%s sec" % (time.clock() - start)
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