反转链表（LeetCode 206. Reverse Linked List）

题目描述

反转一个单链表。

示例:
输入: 1->2->3->4->5->NULL
输出: 5->4->3->2->1->NULL

思路

C++代码

//
// Created by YaMiwan on 2020-05-02.
//

#include 
using namespace std;


// * Definition for singly-linked list.
struct ListNode {
    int val;
    ListNode *next;

    ListNode(int x) : val(x), next(NULL) {}  // 结构体有参默认构造函数传入x值,next指针默认赋为空
};

class Solution {
public:
    ListNode *reverseList(ListNode *head) {  // reverseList:返回链表逆序后的头节点指针，head:链表头节点指针
        ListNode *new_head = NULL;  // 指向新链表头结点的指针
        while (head){
            ListNode *next = head->next;  // 备份head->next，否则会丢失指针
            head->next = new_head;  // 跟新head->next，第一次循环相当于head->next = NULL
            new_head = head;  // 移动new_head
            head = next;  // 遍历原链表
        }
        return new_head;
    }
};

int main(){  // 测试
    ListNode a(1);
    ListNode b(2);
    ListNode c(3);
    ListNode d(4);
    ListNode e(5);
    a.next = &b;  // 将节点简单的链接，进行测试
    b.next = &c;  // 无需构造复杂的链表操作（插入、删除）
    c.next = &d;
    d.next = &e;

    Solution solve;
    ListNode *head = &a;
    cout<<"Before reverse:"<<endl;
    while (head){
        cout<<head->val<<endl;
        head = head->next;
    }

    head = solve.reverseList(&a);  // 逆序
    cout<<"After reverse:"<<endl;
    while (head){
        cout<<head->val<<endl;
        head = head->next;
    }
    return 0;
}

输出

Before reverse:
1
2
3
4
5
After reverse:
5
4
3
2
1