链表逆置

题目链接：反转链表

方法一：递归解法

/**
 * 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) {
        return myReverseList(head);
    }

    ListNode* myReverseList(ListNode* head){
        if(head == NULL || head->next == NULL){
            return head;
        }

        ListNode* newHead = myReverseList(head->next);  //先反转后面的链表
        head->next->next = head;
        head->next= NULL;
        return newHead;
    }
};

方法二：借助于辅助指针

/**
 * 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) {
        ListNode* last = NULL;
        ListNode* p,*q;
        p = head;
        while(p){
            q = p->next;
            p->next = last;
            last = p;
            if(q == NULL){
                head = p;
            }
            p = q;
        }
        return head;
    }
};

链表逆置

原文：https://www.cnblogs.com/ManOK/p/13723525.html