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面试题24. 反转链表

题目描述

定义一个函数，输入一个链表的头节点，反转该链表并输出反转后链表的头节点。

示例:

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

限制：

• 0 <= 节点个数 <= 5000

解法

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:
        if not head:
            return head
        dummy = ListNode(0)
        p = head
        while p:
            q = p.next
            p.next = dummy.next
            dummy.next = p
            p = q
        return dummy.next

Java

/**
 * 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) {
            return head;
        }
        ListNode dummy = new ListNode(0);
        ListNode p = head;
        while (p != null) {
            ListNode q = p.next;
            p.next = dummy.next;
            dummy.next = p;
            p = q;
        }
        return dummy.next;
    }
}

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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