递归_反转链表_C++

一.题目解析

算法解析:1.循环

三指针法:prev,cur,next,然后cur指针指向prev实现反转,注意指针前进的顺序

代码实现

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        ListNode* prev=nullptr;
        ListNode* cur=head;
        
        while(cur!=nullptr)
        {
            ListNode* next=cur->next;
            cur->next=prev;
            prev=cur;
            cur=next;
        }
        return prev;
    }
};

算法解析2:递归

思路1:

思路2:

递归代码实现

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        if(head==nullptr||head->next==nullptr)return head;//边界情况

        ListNode*newhead= reverseList(head->next);//深度优先遍历
        head->next->next=head;
        head->next=nullptr;
        return newhead;
    }
};