LeetCode-Reverse Linked List I & II

两个链表反转问题。

206. Reverse Linked List

Reverse a singly linked list.

Hint:

A linked list can be reversed either iteratively or recursively. Could you implement both?

Solution

iteratively

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

class Solution(object):
    def reverseList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        rev = None
        while head:
            head.next,rev,head = rev,head,head.next
        return rev

recursively

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

class Solution(object):
    def reverseList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        return self.reverse_recursively(head, None)
        
    def reverse_recursively(self, head, new_head):
        if not head:
            return new_head
        nxt = head.next
        head.next = new_head
        return self.reverse_recursively(nxt, head)

92. 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.

Solution

# 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
        """
        if not head:
            return None
        dummy = ListNode(None)
        dummy.next = head
        pre = dummy
        for i in range(m-1):
            pre = pre.next
        start = pre.next
        then = start.next
        for i in range(n-m):  # n到m之间有n-m+1个节点，需要反转n-m次
            # 每次不断的把then节点反转到pre节点的后面，然后then节点后移一个，直到反转完毕
            start.next = then.next
            then.next = pre.next
            pre.next = then
            then = start.next
        return dummy.next

更加直观的解法：

# 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 = ListNode(None)
        dummy.next = head
        whead = head  # 窗口开始位置
        wtail = head  # 窗口结束位置
        ohead = dummy  # 窗口的前一个位置
        for i in range(n-m):  # 初始化窗口长度
            wtail = wtail.next
        for i in range(m-1):  # 移动窗口到位置m
            ohead, whead, wtail = whead, whead.next, wtail.next
        otail = wtail.next  # 窗口结束的后一个位置
        wtail.next = None
        revhead, revtail = self.reverse(whead)  # 反转窗口内部链表
        ohead.next, revtail.next = revhead, otail  # 将反转后的窗口与未反转部分连接起来
        return dummy.next
        
    def reverse(self, head):  # 反转链表
        pre = None
        cur = tail = head
        while cur:
            cur.next, pre, cur = pre, cur, cur.next
        return pre, tail