[leetcode] Reverse Linked List ii

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.

解題思路：翻轉鏈表的題目。請用積木化思維(Building block): 

這裏必須的積木：鏈表翻轉操作：

current.next, prev, current = prev, current, current.next

我自己的代碼很繁瑣，在網上看到的代碼非常簡潔，學習了。。

轉自：http://www.cnblogs.com/asrman/p/3971933.html

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

class 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