[LeetCode] Reverse Linked List(递归与非递归反转链表)

来源：互联网发布：简历封面知乎编辑：程序博客网时间：2024/06/14 03:57

Reverse a singly linked list.

解题思路

对于非递归实现，思路是依次将从第二个结点到最后一个结点的后继设为头结点，然后将该节点设为头结点（需记住将原头结点的后继设为空）。
对于递归实现，首先反转从第二个结点到最后一个结点的链表，然后再将头结点放到已反转链表的最后，函数返回新链表的头结点。

非递归实现代码1[C++]

//Runtime:10 msclass Solution {public:    ListNode* reverseList(ListNode* head) {        if (head == NULL) return NULL;        ListNode *pre = head;        ListNode *cur = head->next;        while (cur != NULL)        {            pre->next = cur->next;            cur->next = head;            head = cur;            cur = pre->next;        }        return head;    }};

非递归实现代码2[C++]

//Runtime:22 msclass Solution{public:    ListNode* reverseList(ListNode* head){        if (head == NULL) return NULL;        ListNode *cur = head->next;        head->next = NULL;        while (cur != NULL)        {            ListNode *temp = cur->next;            cur->next = head;            head = cur;            cur = temp;        }        return head;    }};

非递归实现代码3[Java]

//Runtime:276 mspublic class Solution {    public ListNode reverseList(ListNode head) {        if (head == null) {            return head;        }        ListNode cur = head.next;        ListNode pre = head;        while (cur != null){            pre.next = cur.next;            cur.next = head;            head = cur;            cur = pre.next;        }        return head;    }}

非递归实现代码4[Python]

#Runtime: 63 ms# Definition for singly-linked list.class ListNode:    def __init__(self, x):        self.val = x        self.next = Noneclass Solution:    # @param {ListNode} head    # @return {ListNode}    def reverseList(self, head):        if head == None:            return head        pre = head        cur = head.next        while cur != None:            pre.next = cur.next            cur.next = head            head = cur            cur = pre.next        return heads = Solution()head = ListNode(0);cur = headfor i in range(1, 10):    node = ListNode(i)    cur.next = node    cur = nodehead = s.reverseList(head);while(head != None):    print(head.val, end=' ')    head = head.next

递归实现代码[C++]

//Runtime:10 msclass Solution{public:    ListNode* reverseList(ListNode* head){        //此处的条件不能写成if(head == NULL)        if (head == NULL || head->next == NULL) return head;        ListNode *newhead = reverseList(head->next);        head->next->next = head;        head->next = NULL;        return newhead;    }};

C++测试代码如下：

void printList(ListNode *head){    while(head != NULL)    {        cout<<head->val<<" ";        head = head->next;    }    cout<<endl;}void dropList(ListNode *head){    if (head == NULL) return;    ListNode *temp;    while (head != NULL)    {        temp = head->next;        delete head;        head = temp;    }}int main(){    ListNode *head = new ListNode(0);    ListNode *cur = head;    for (int i = 0; i < 10; i++)    {        ListNode *newnode = new ListNode(floor(rand()%100));        cur->next = newnode;        cur = newnode;    }    printList(head);    Solution s;    head = s.reverseList(head);    printList(head);    dropList(head);}

Java测试代码如下：

public class AA {    public static void main(String[] args) {        // TODO Auto-generated method stub        ListNode head = new ListNode(0);        ListNode cur = head;        for(int i = 1; i < 10; i++){            ListNode node = new ListNode(i);            cur.next = node;            cur = node;        }        Solution s = new Solution();        head = s.reverseList(head);        print(head);    }    static void print(ListNode head){        while(head != null){            System.out.println(head.val);            head = head.next;        }    }}

2 0