題目來自Leetcode

206. Reverse Linked List

Reverse a singly linked list.

Hint:

A linked list can be reversed eitheriteratively or recursively. Could you implement both?

classSolution {
public:
    ListNode* reverseList(ListNode* head) {
        //iterative version using 3 pointers
        ListNode* p,*q,*r;
        //empty list
        if(!head)
            return NULL;
        p=head;
        q=p->next;
        //only one node
        if(q==NULL)
            return p;
        else{
            if(q->next!=NULL)
                r=q->next;
            else
                r=NULL;
        }
        p->next=NULL;
        q->next=p;
        while(r){
            p=q;
            q=r;
            r=q->next;
            q->next=p;
        }
        return q;
    }
};
 

下面是用遞迴的解法：

classSolution {
public:
    ListNode* reverseList(ListNode* head) {
        if(head==NULL || head->next==NULL)
        returnhead;
       ListNode*smallo=reverseList(head->next);
        ListNode*tail=head->next;
        tail->next=head;
        head->next=NULL;
        return smallo;
    }
};
 

類似地。。。

classSolution {
public:  
    ListNode* reverseList(ListNode* head) {
        if (!head || !(head -> next)) returnhead;
        ListNode* node = reverseList(head ->next);
        head -> next -> next = head;
        head -> next = NULL;
        return node;
    }
};