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剑指 Offer II 024. 反转链表

题目描述

给定单链表的头节点 head ，请反转链表，并返回反转后的链表的头节点。

 

示例 1：

输入：head = [1,2,3,4,5]
输出：[5,4,3,2,1]

示例 2：

输入：head = [1,2]
输出：[2,1]

示例 3：

输入：head = []
输出：[]

 

提示：

  • 链表中节点的数目范围是 [0, 5000]
  • -5000 <= Node.val <= 5000

 

进阶：链表可以选用迭代或递归方式完成反转。你能否用两种方法解决这道题？

 

注意：本题与主站 206 题相同： https://leetcode.cn/problems/reverse-linked-list/

解法

方法一

思考

就地反转单链表。额外数组记下结点再倒序重连需要线性空间。

迭代时保存后继，把当前 \(\textit{next}\) 改向前驱，再整体前移。遍历结束后前驱即为新头。

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# 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

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/**
 * 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;
    }
}

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/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        ListNode* pre = nullptr;
        ListNode* p = head;
        while (p) {
            ListNode* q = p->next;
            p->next = pre;
            pre = p;
            p = q;
        }
        return pre;
    }
};

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/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func reverseList(head *ListNode) *ListNode {
    var pre *ListNode
    for p := head; p != nil; {
        q := p.Next
        p.Next = pre
        pre = p
        p = q
    }
    return pre
}

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/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} head
 * @return {ListNode}
 */
var reverseList = function (head) {
    let pre = null;
    for (let p = head; p;) {
        let q = p.next;
        p.next = pre;
        pre = p;
        p = q;
    }
    return pre;
};

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/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int val=0, ListNode next=null) {
 *         this.val = val;
 *         this.next = next;
 *     }
 * }
 */
public class Solution {
    public ListNode ReverseList(ListNode head) {
        ListNode pre = null;
        for (ListNode p = head; p != null;) {
            ListNode t = p.next;
            p.next = pre;
            pre = p;
            p = t;
        }
        return pre;
    }
}

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/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public var val: Int
 *     public var next: ListNode?
 *     public init(_ val: Int) {
 *         self.val = val
 *         self.next = nil
 *     }
 * }
 */

 class Solution {
    func reverseList(_ head: ListNode?) -> ListNode? {
        var prev: ListNode? = nil
        var current = head

        while current != nil {
            let next = current?.next
            current?.next = prev
            prev = current
            current = next
        }

        return prev
    }
}

方法二

思考

方法一已在原地完成反转。递归先把后半段反转，再把当前结点接到后半段尾部并断开原后继，语义更贴近「先处理子问题」。

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/**
 * 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;
    }
}

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