[LeetCode] 206. Reverse Linked List 反向链表

Reverse a singly linked list.

Hint:

A linked list can be reversed either iteratively or recursively. Could you implement both?

反向链表，分别用递归和迭代方式实现。

递归Iteration：

新建一个node(value=任意值, next = None)， 用一个变量 next 记录head.next，head.next指向新node.next，新 node.next 指向head，next 进入下一个循环，重复操作，直到结束。

迭代Recursion：

假设有链表1 -> 2 -> 3 -> 4 -> 5要进行反转，第一层先把1 -> 2 -> 3 -> 4 -> 5和None传入递归函数(head, newhead)，先记录head的next(2 -> 3 -> 4 -> 5)，head(1)指向newhead(None)变成(1->None)，然后把(2 -> 3 -> 4 -> 5,  1 -> None)递归传入下一层，记录head的next(3 -> 4 -> 5)，head(2)指向newhead(1->None)变成(2 -> 1->None)，然后把(3 -> 4 -> 5,  2 -> 1 -> None)递归传入下一层，直到最后head=None时返回newhead，此时 newhead = 5 -> 4 -> 3 -> 2 -> 1 -> None

Java: Iterative solution

public ListNode reverseList(ListNode head) {
    ListNode newHead = null;
    while (head != null) {
        ListNode next = head.next;
        head.next = newHead;
        newHead = head;
        head = next;
    }
    return newHead;
}

Java: Recursive solution

public ListNode reverseList(ListNode head) {
    return reverseListInt(head, null);
}

private ListNode reverseListInt(ListNode head, ListNode newHead) {
    if (head == null)
        return newHead;
    ListNode next = head.next;
    head.next = newHead;
    return reverseListInt(next, head);
}　　

Python: Iteration

def reverseList(self, head):
    prev = None
    while head:
        curr = head
        head = head.next
        curr.next = prev
        prev = curr
    return prev

Python: Iteration

def reverseList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        cur, prev = head, None
        while cur:
            cur.next, prev, cur = prev, cur, cur.next
        return prev

Python: Iteration

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

    def __repr__(self):
        if self:
            return "{} -> {}".format(self.val, repr(self.next))

class Solution:
    def reverseList(self, head):
        dummy = ListNode(0)
        while head:
            next = head.next
            head.next = dummy.next
            dummy.next = head
            head = next
        return dummy.next

Python: Recrusion

class Solution(object):
    def reverseList(self, head, prev=None):
        if not head:
          return prev

        curr, head.next = head.next, prev
        return self.reverseList(curr, head)　　

Python: Recursion

class Solution:
    def reverseList(self, head):
        return self.doReverse(head, None)
    def doReverse(self, head, newHead):
        if head is None:
            return newHead
        next = head.next
        head.next = newHead
        return self.doReverse(next, head)

Python: Recursion, wo

class Solution(object):
    def reverseList(self, head):
        dummy = ListNode(0)
        cur = self.recur(head, dummy)
        return cur.next

    def recur(self, head1, head2):
        if not head1:
            return head2

        next_node = head1.next
        head1.next = head2.next
        head2.next = head1
        return self.recur(next_node, head2)　　

C++: Iteration

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        auto dummy = ListNode{0};

        while (head) {
            auto tmp = head->next;
            head->next = dummy.next;
            dummy.next = head;
            head = tmp;
        }

        return dummy.next;
    }
};

C++: Recursion

class Solution {
public:
    /**
     * @param head: The first node of linked list.
     * @return: The new head of reversed linked list.
     */
    ListNode *reverse(ListNode *head) {
        if (!head || !head->next) return head;
        ListNode *t = head;
        head = reverse(t->next);
        t->next->next = t;
        t->next = NULL;
        return head;
    }
};

　　　

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