Reverse Linked List

这道题是把链表倒过来......活久见递归自己搞对了......：

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

或者不用递归，用迭代：

/**
 * 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 *cur=head;
        ListNode *pre=NULL;
        while(cur!=NULL)
        {
            ListNode *temp=cur->next;
            cur->next=pre;
            pre=cur;
            cur=temp;
        }
        return pre;
    }
};