leetcode 206-- flipped list

leetcode 206-- flipped list

Subject description:

Reverse a singly linked list.

Example:

Input: 1-> 2-> 3-> 4- > 5-> NULL
Output: 5-> 4-> 3-> 2- > 1-> NULL
Advanced:
You may iteratively or recursively inverted list. Can you solve this question in two ways?

Source: stay button (LeetCode)
link: https: //leetcode-cn.com/problems/reverse-linked-list
copyrighted by deduction from all networks. Commercial reprint please contact the authorized official, non-commercial reprint please indicate the source.

Submit a: Iteration

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
//迭代法翻转链表
//将首节点pMove的后面节点nextNode拿到头部 head
//重复这一过程，直到原来的首结点pMove移动到尾节点
class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        if(head==nullptr||head->next==nullptr) return head;
        
        ListNode* pMove = head;
        while(pMove->next){//当pMove没有到达尾节点时迭代进行(循环)
            ListNode* nextNode = pMove->next;    //将pMove->next得到    
            //将nextNode拿到头结点
            pMove->next = nextNode->next;
            nextNode->next = head;
            head = nextNode;
        }
        return head;
    }
};

Submit two: this iteration is also quite interesting (excerpt from leetcode official problem-solving)

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
 // 4 -> 3 -> 7 -> 9 -> nullptr
 //nullptr <- 9 <- 7 <- 3 <- 4
 //将当前结点curNode的指向由指向右边改为指向左边 
class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        ListNode* frontNode = nullptr;
        ListNode* curNode = head;
        while(curNode){
            ListNode* nextNode = curNode->next;
            curNode->next = frontNode;
            frontNode = curNode;
            curNode = nextNode;
        }
        return frontNode;
    }
};

Submission: the magic of recursion

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}![在这里插入图片描述](https://img-blog.csdnimg.cn/20200210161921279.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80NDc5NTgzOQ==,size_16,color_FFFFFF,t_70)
 * };
 */
 //神奇的递归
class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        if(head==nullptr||head->next==nullptr) return head;
        ListNode* headNode = reverseList(head->next);
        head->next->next = head;
        head->next =nullptr;
        return headNode;
    }
};

Problem Solving: leetcode official problem-solving: