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[LeetCode] Reverse Linked List II @ Python
原题地址：https://oj.leetcode.com/problems/reverse-linked-list-ii/
题意：
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.
解题思路：翻转链表的题目。请用积木化思维(Building block):
这里必须的积木：链表翻转操作：
current.next, prev, current = prev, current, current.next
代码：
# Definition for singly-linked list.# class ListNode:# def __init__(self, x):# self.val = x# self.next = Noneclass Solution: # @param head, a ListNode # @param m, an integer # @param n, an integer # @return a ListNode def reverseBetween(self, head, m, n): diff, dummy = n - m, ListNode(0) dummy.next = head prev, current = dummy, dummy.next while current and m >1: prev, current = current, current.next m -= 1 last_swapped, first_swapped = prev, current while current and diff >= 0: current.next, prev, current = prev, current, current.next diff -= 1 last_swapped.next, first_swapped.next = prev, current return dummy.next # Reverse partial Linked List # Refer Figure1: http://images.cnitblog.com/i/546654/201404/072244468407048.jpg # Refer: http://www.cnblogs.com/4everlove/p/3651002.html # Refer: http://stackoverflow.com/questions/21529359/reversing-a-linked-list-in-python # 1 --> 2 --> 3 --> 4 --> 5 # # first_swapped.next = current # __________________ # ^ | # | V # 1 2 <-- 3 <-- 4 5 # | ^ # V ________________| # last_unswapped.next=prev
[LeetCode] Reverse Linked List II @ Python
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