leetcode 简单题 206 反转链表

一、题目：
反转一个单链表。
示例:
输入: 1->2->3->4->5->NULL
输出: 5->4->3->2->1->NULL

二、解法：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) {
        if(head == NULL)
            return NULL;
        ListNode* present = head->next;//从第二个节点开始和head换
        ListNode* prevpresent = head;//present的前一个
        while(present!=NULL){
            prevpresent->next = present->next;
            present->next = head;
            head = present;
            present = prevpresent->next;
        }
        return head;
    }
};```

三、解法：递归：

```cpp
/**
 * 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)
            return head;
        if(head->next==NULL)//last node
            return head;
        ListNode* newhead = reverseList(head->next);
        ListNode* temp = newhead;
        while(temp->next!=NULL)
            temp = temp->next;
        temp->next = head;
        temp->next->next = NULL;
        return newhead;
    }
};```

递归迭代区别：
详见：https://blog.csdn.net/u011514810/article/details/52749183