反转链表

leetcode原题： 206. 反转链表

题目描述

反转一个单链表。

示例:

输入: 1->2->3->4->5->NULL
输出: 5->4->3->2->1->NULL

进阶:
你可以迭代或递归地反转链表。你能否用两种方法解决这道题？

分析

迭代思路：

把第一个Node置为null，然后在迭代过程中总是把后一个Node放到链表头部。

递归思路：

第一种方式：类似于上面的迭代思路，把第一个Node置为null，相邻两个Node反转后，继续取后一个Node参与反转。

第二种方式：先反转head.next，然后再把head放到链表尾部。反转head.next正好是一个递归的过程。

实现

Java

迭代版本：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode reverseList(ListNode head) {
        ListNode pre = null;
        ListNode cur = head;
        while (cur != null) {
            ListNode next = cur.next;
            cur.next = pre;
            pre = cur;
            cur = next;
        }
        return pre;
    }
}

递归版本(一)：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode reverseList(ListNode head) {
        return reverse(null, head);
    }

    public ListNode reverse(ListNode pre, ListNode cur) {
        if (cur == null) {
            return pre;
        }
        ListNode next = cur.next;
        cur.next = pre;
        return reverse(cur, next);
    }
}

递归版本(二)：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode reverseList(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }

        ListNode res = reverseList(head.next);
        head.next.next = head;
        head.next = null;

        return res;
    }
}

Python3

迭代版本：

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reverseList(self, head: ListNode) -> ListNode:
        pre, cur = None, head
        while cur:
            cur.next, pre, cur = pre, cur, cur.next
        return pre

递归版本(一)：

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reverseList(self, head: ListNode) -> ListNode:
        
        def reverse(pre: ListNode, cur: ListNode) -> ListNode:
            if not cur:
                return pre
            next_node = cur.next
            cur.next = pre
            return reverse(cur, next_node)

        return reverse(None, head)

递归版本(二)：

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reverseList(self, head: ListNode) -> ListNode:
        if not head or not head.next:
            return head
        res = self.reverseList(head.next)
        head.next.next, head.next = head, None
        return res