LeetCode 92 Reverse Linked List II (Python详解及实现)

【题目】

Reverse a linked list from position m to n.Do it in-place and in one-pass.

 

For example:

Given 1->2->3->4->5->NULL, m= 2 and n = 4,

 

return 1->4->3->2->5->NULL.

 

Note:

Given m, n satisfy the following condition:

1 ≤ m ≤ n ≤ length of list.

 

 

【思路】

定义指针关注是否到m和n的位置，到达后m后将链表断开成左右两部分A、B，继续移动到达n时将链表分成两部分B、C，这样可以提取出m,n段即B部分，然后将B部分反转后，与AC重新连接即可。

例1->2->3->4->5->NULL过程：

l  初始状态

#     0 ----> 1 ----> 2 ----> 3 ----> 4 ----> 5

#     prev  curr

l  第一个while循环（m=2,n=4）执行完状态：

       #      0 ----> 1 ----> 2 ---->3 ----> 4 ----> 5

       #          prev  curr

l  执行last_unswapped,first_swapped = prev, curr，开始断开的结点m前一个点

         #      0 -------> 1 ----------------------->2 ------------------------> 3 ----> 4 ----> 5

    #          last_unswapped   first_swapped

l  第二个while循环（m=2,n=4）执行状态：

        # diff=2 #  1 <-- 2     3 ----> 4 ---->5

       #                  prev curr

       # diff=1 #   1 <-- 2 <--3     4 ----> 5

       #                       prev  curr

       # diff=0 #   1 <-- 2 <-- 3<-- 4     5

       #                           prev  curr

l  执行last_unswapped.next= prev，first_swapped.next  = curr，结束断开的结点n后的点。执行上main两条语句后状态：



【Python实现】

# -*- coding: utf-8 -*-
"""
Created on Thu Aug 10 10:18:04 2017

@author: Administrator
"""
# Definition for singly-linked list.
class ListNode(object):
def __init__(self, x):
self.val = x
self.next = None

class Solution(object):
def reverseBetween(self, head, m, n):
"""
:type head: ListNode
:type m: int
:type n: int
:rtype: ListNode
"""
dummy, partial_len = ListNode(0), n - m
dummy.next = head
prev, curr = dummy, dummy.next
#取出mn之间的结点，并反转
while m > 1:#pre是开始反转的结点
prev, curr = curr, curr.next
m -= 1
last_unswapped, first_swapped = prev, curr
while curr and partial_len >= 0:#pre是结束反转的结点
curr.next, prev, curr = prev, curr, curr.next
partial_len -= 1
#重新连接结点
last_unswapped.next, first_swapped.next = prev, curr
return dummy.next

if __name__ == '__main__':
S = Solution()
l1 = ListNode(1)
l2 = ListNode(2)
l3 = ListNode(3)
l4 = ListNode(4)
l5 = ListNode(5)
head = l1
l1.next = l2
l2.next = l3
l3.next = l4
l4.next = l5
l5.next = None
S.partition(head,2,4)