LeetCode Reverse Linked List II

LeetCode Reverse Linked List II
Reverse Linked List II

Original question

When only one traversal is performed and no additional space is requested, the elements m to n in a linked list are flipped over.

Note:

M and n meet the following conditions: 1 ≤ m ≤ n ≤ length of the linked list

Example:

Input: 1-> 2-> 3-> 4-> 5-> NULL, m = 2, n = 4

Output: 1-> 4-> 3-> 2-> 5-> NULL

Solutions

You can use the Reverse Linked List to flip the Linked List. Look, you can flip a section first, and then connect it to the front and back linked list. Because the first node may have to be flipped over, a false header node is added for consistency.

AC Source Code

# Definition for singly-linked list.class ListNode(object):    def __init__(self, x):        self.val = x        self.next = None    def to_list(self):        return [self.val] + self.next.to_list() if self.next else [self.val]class Solution(object):    def reverseBetween(self, head, m, n):        """        :type head: ListNode        :type m: int        :type n: int        :rtype: ListNode        """        dummy = ListNode(-1)        dummy.next = head        node = dummy        for __ in range(m - 1):            node = node.next        prev = node.next        curr = prev.next        for __ in range(n - m):            next = curr.next            curr.next = prev            prev = curr            curr = next        node.next.next = curr        node.next = prev        return dummy.nextif __name__ == "__main__":    n1 = ListNode(1)    n2 = ListNode(2)    n3 = ListNode(3)    n4 = ListNode(4)    n5 = ListNode(5)    n1.next = n2    n2.next = n3    n3.next = n4    n4.next = n5    r = Solution().reverseBetween(n1, 2, 4)    assert r.to_list() == [1, 4, 3, 2, 5]