Reverse Linked List

题目：

单链表逆置，必须掌握

Reverse a singly linked list.

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Hint:

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

解析：

迭代

排除没有节点和单个节点的情况后，用a,b,c标记三个相邻链表，让b指向a然后在将三个节点顺次移动。注意要判断c是否存在，因为c若已经为null的话c->next会引发时间超出的错误

/**
 * 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* a = head;
        ListNode* b = a->next;
        ListNode* c = b->next;
        a->next = NULL;
        while(b != NULL)
        {
            b->next = a;
            a = b;
            b = c;
            if(c != NULL)
                c = c->next;
        }
        return a;
    }
};

递归

递归比较难想啊：

class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        if (head == NULL || head->next == NULL) {
            return head;
        }
        ListNode* root = reverseList(head->next);
        head->next->next = head;
        head->next = NULL;
        return root;
    }
};