LeetCode: Reverse Nodes in kGroup

从这题和上一题可以总结出反转链表的经验，需要有5个指针：end, q, p, pPre, pNext. p和pPre进行方向转置后p和pPre向后移，pNext用来记录转置前p的后一个，q用来记录转置串之前的node，end记录转置串最开始的node。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *reverseKGroup(ListNode *head, int k) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        ListNode *pPre, *end, *p, *q;
        p = head;
        q = NULL;
        bool flag = true;
        while (flag) {
            end = pPre = p;
            ListNode *tmp = pPre;
            for (int i = 0; i < k; i++) {
                if (tmp) tmp = tmp->next;
                else {
                    flag = false;
                    break;
                }
            }
            if (!flag) break;
            p = p->next;
            for (int i = 0; i < k-1; i++) {
                ListNode *pNext = p->next;
                p->next = pPre;
                pPre = p;
                p = pNext;
            }
            end->next = p;
            if (!q) head = pPre;
            else q->next = pPre;
            q = end;
        }
        return head;
    }
};

 recursive的方法更好理解

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *reverseKGroup(ListNode *head, int k) {
        if (!head) return NULL;
        ListNode *p = head;
        int len = 0;
        for (; p && len < k; p = p->next) len++;
        if (len < k) return head;
        ListNode *pre = NULL;
        p = head;
        for (int i = 0; i < k; ++i) {
            ListNode *pnext = p->next;
            p->next = pre;
            pre = p;
            p = pnext;
        }
        head->next = reverseKGroup(p, k);
        return pre;
    }
};

 C#

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode ReverseKGroup(ListNode head, int k) {
        if (head == null) return null;
        ListNode p = head;
        int len = 0;
        for (; p != null && len < k; p = p.next) len++;
        if (len < k) return head;
        ListNode pre = null;
        p = head;
        for (int i = 0; i < k; i++) {
            ListNode pNext = p.next;
            p.next = pre;
            pre = p;
            p = pNext;
        }
        head.next = ReverseKGroup(p, k);
        return pre;
    }
}