LeetCode92 Reverse Linked List II

题目链接:

https://leetcode.com/problems/reverse-linked-list-ii/

题目描述:

给一个链表，例如1->2->3->4->5->NULL。

m=2,n=4，将m到n之间的链表反向。1->4->3->2->5->NULL

分析:

和链表反向没啥区别，就是多了一些控制而已。只是在写reverseN函数时，开始是想用指针的引用来写的，QAQ然而发现，我还是写的磕磕绊绊，之后补上。

代码：

class Solution {
public:
    ListNode* reverseN(ListNode* &head, ListNode* ptr, int cnt, int n){
        if (cnt == n){
            head = ptr;
            return head;
        }
        ListNode* pNext=reverseN(head, ptr->next, cnt + 1, n);
        ListNode* tmp = pNext->next;
        pNext->next = ptr;
        ptr->next = tmp;
        return ptr;
   }
    ListNode* reverseBetween(ListNode* head, int m, int n) {
        if (m == n){
            return head;
        }
        ListNode* pHead = (ListNode*)malloc(sizeof(ListNode));
        pHead->next = head;
        int cnt = 1;
        ListNode* pre = pHead;
        ListNode* cur = head;
        while (cur != NULL){
            if (cnt == m){
                reverseN(cur, cur, cnt, n);
                pre->next = cur;
                break;
            }
            cnt++;
            pre = cur;
            cur = cur->next;
        }
        return pHead->next;
    }
};