206. 反转链表 (opens new window)

English Version (opens new window)

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？

解法

定义指针 p、q 分别指向头节点和下一个节点，pre 指向头节点的前一个节点。

遍历链表，改变指针 p 指向的节点的指向，将其指向 pre 指针指向的节点，即 p.next = pre。然后 pre 指针指向 p，p、q 指针往前走。

当遍历结束后，返回 pre 指针即可。

Python3

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

class Solution:
    def reverseList(self, head: ListNode) -> ListNode:
        pre, p = None, head
        while p:
            q = p.next
            p.next = pre
            pre = p
            p = q
        return pre

Java

迭代版本：

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

递归版本：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
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;
    }
}

JavaScript

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */
/**
 * @param {ListNode} head
 * @return {ListNode}
 */
var reverseList = function (head) {
  let node = head;
  let pre = null;
  while (node) {
    let cur = node;
    node = cur.next;
    cur.next = pre;
    pre = cur;
  }
  return pre;
};

Go

func reverseList(head *ListNode) *ListNode {
    if head == nil ||head.Next == nil {
        return head
    }
    dummyHead := &ListNode{}
    cur := head
    for cur != nil {
        tmp := cur.Next
        cur.Next = dummyHead.Next
        dummyHead.Next = cur
        cur = tmp
    }
    return dummyHead.Next
}

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