反转链表--206，92 迭代 递归

在想这迭代和递归到底是怎么变换的，必须要理清楚指针。

迭代写法
图像理解：

206:

/**
 * 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* dummy = new ListNode(-1);
        dummy->next = head;
        ListNode *tail = head;
        while(tail && tail->next)
        {
            ListNode* cur = tail->next;
            ListNode* nxt = cur->next;
            cur->next = dummy->next;
            dummy->next = cur;
            tail->next = nxt;
        }
        return dummy->next;
    }
};

92:

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* reverseBetween(ListNode* head, int m, int n) {
        ListNode* dummy = new ListNode(-1);
        dummy->next = head;
        ListNode* front = dummy;
        for(int i = 1; i < m; i++)
            front = front->next;
        ListNode* tail = front->next;
        for(int i = m + 1; i<= n; i++)
        {
            ListNode* cur = tail->next;
            ListNode* nxt = cur->next;
            cur->next = front->next;
            front->next = cur;
            tail->next = nxt;
        }
        return dummy->next;
    }
};

递归
图解：

206：

/**
 * 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 || !head->next) return head;
        ListNode* p = reverseList(head->next);
        head->next->next = head;
        head->next = nullptr;
        return p;
    }
};

92：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
    ListNode* tail;
    ListNode* dfs(ListNode* head, int cnt)
    {
        tail = head->next;
        if(cnt == 1) return head;
        ListNode* p = dfs(head->next, cnt - 1);
        head->next->next = head;
        head->next = tail;
        return p;
    }
public:
    ListNode* reverseBetween(ListNode* head, int m, int n) {
        if(!head) return nullptr;
        ListNode* dummy = new ListNode(-1);
        dummy->next = head;
        ListNode* front = dummy;
        for(int i = 1; i < m; i++)
            front = front->next;
        front->next = dfs(front->next, n - m + 1);
        return dummy->next;
    }
};