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Reverse Linked List

Source

Reverse a linked list.

Example
For linked list 1->2->3, the reversed linked list is 3->2->1

Challenge
Reverse it in-place and in one-pass

题解1 - 非递归

联想到同样也可能需要翻转的数组，在数组中由于可以利用下标随机访问，翻转时使用下标即可完成。而在单向链表中，仅仅只知道头节点，而且只能单向往前走，故需另寻出路。分析由1->2->3变为3->2->1的过程，由于是单向链表，故只能由1开始遍历，1和2最开始的位置是1->2，最后变为2->1，故从这里开始寻找突破口，探讨如何交换1和2的节点。

temp = head->next;
head->next = prev;
prev = head;
head = temp;

要点在于维护两个指针变量prev和head, 翻转相邻两个节点之前保存下一节点的值，分析如下图所示：

Reverse Linked List

  1. 保存head下一节点
  2. 将head所指向的下一节点改为prev
  3. 将prev替换为head，波浪式前进
  4. 将第一步保存的下一节点替换为head，用于下一次循环

Python

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

class Solution:
    # @param {ListNode} head
    # @return {ListNode}
    def reverseList(self, head):
        prev = None
        curr = head
        while curr is not None:
            temp = curr.next
            curr.next = prev
            prev = curr
            curr = temp
        # fix head
        head = prev

        return head

C++

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* reverse(ListNode* head) {
        ListNode *prev = NULL;
        ListNode *curr = head;
        while (curr != NULL) {
            ListNode *temp = curr->next;
            curr->next = prev;
            prev = curr;
            curr = temp;
        }
        // fix head
        head = prev;

        return head;
    }
};

Java

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode reverseList(ListNode head) {
        ListNode prev = null;
        ListNode curr = head;
        while (curr != null) {
            ListNode temp = curr.next;
            curr.next = prev;
            prev = curr;
            curr = temp;
        }
        // fix head
        head = prev;

        return head;
    }
}

源码分析

题解中基本分析完毕，代码中的prev赋值比较精炼，值得借鉴。

复杂度分析

遍历一次链表，时间复杂度为 O(n)O(n), 使用了辅助变量，空间复杂度 O(1)O(1).

题解2 - 递归

递归的终止步分三种情况讨论：

  1. 原链表为空，直接返回空链表即可。
  2. 原链表仅有一个元素，返回该元素。
  3. 原链表有两个以上元素，由于是单链表，故翻转需要自尾部向首部逆推。

由尾部向首部逆推时大致步骤为先翻转当前节点和下一节点，然后将当前节点指向的下一节点置空(否则会出现死循环和新生成的链表尾节点不指向空)，如此递归到头节点为止。新链表的头节点在整个递归过程中一直没有变化，逐层向上返回。

Python

"""
Definition of ListNode

class ListNode(object):

    def __init__(self, val, next=None):
        self.val = val
        self.next = next
"""
class Solution:
    """
    @param head: The first node of the linked list.
    @return: You should return the head of the reversed linked list.
                  Reverse it in-place.
    """
    def reverse(self, head):
        # case1: empty list
        if head is None:
            return head
        # case2: only one element list
        if head.next is None:
            return head
        # case3: reverse from the rest after head
        newHead = self.reverse(head.next)
        # reverse between head and head->next
        head.next.next = head
        # unlink list from the rest
        head.next = None

        return newHead

C++

/**
 * Definition of ListNode
 *
 * class ListNode {
 * public:
 *     int val;
 *     ListNode *next;
 *
 *     ListNode(int val) {
 *         this->val = val;
 *         this->next = NULL;
 *     }
 * }
 */
class Solution {
public:
    /**
     * @param head: The first node of linked list.
     * @return: The new head of reversed linked list.
     */
    ListNode *reverse(ListNode *head) {
        // case1: empty list
        if (head == NULL) return head;
        // case2: only one element list
        if (head->next == NULL) return head;
        // case3: reverse from the rest after head
        ListNode *newHead = reverse(head->next);
        // reverse between head and head->next
        head->next->next = head;
        // unlink list from the rest
        head->next = NULL;

        return newHead;
    }
};

Java

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode reverse(ListNode head) {
        // case1: empty list
        if (head == null) return head;
        // case2: only one element list
        if (head.next == null) return head;
        // case3: reverse from the rest after head
        ListNode newHead = reverse(head.next);
        // reverse between head and head->next
        head.next.next = head;
        // unlink list from the rest
        head.next = null;

        return newHead;
    }
}

源码分析

case1 和 case2 可以合在一起考虑，case3 返回的为新链表的头节点，整个递归过程中保持不变。

复杂度分析

递归嵌套层数为 O(n)O(n), 时间复杂度为 O(n)O(n), 空间(不含栈空间)复杂度为 O(1)O(1).

Reference