[leetcode 25]Reverse Nodes in k-Group

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本文介绍了一种算法，该算法可以将链表按K个一组进行反转，并保持剩余节点不变。通过对链表进行递归操作，实现了仅使用常量内存完成任务的目标。文章提供了完整的C++实现代码。

Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.

If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.

You may not alter the values in the nodes, only nodes itself may be changed.

Only constant memory is allowed.

For example,
Given this linked list: 1->2->3->4->5

For k = 2, you should return: 2->1->4->3->5

For k = 3, you should return: 3->2->1->4->5

/**
 * 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 == NULL || head->next == NULL || k < 2) {
            return head;
        }
        ListNode dummy(-1);
        dummy.next = head;
        for (ListNode *prev = &dummy, *end = head; end; end = prev->next) {
            for (int i = 1; i < k && end; i++) {
                end = end->next;
            }
            if (end == NULL) {
                break;
            }
            prev = recur(prev, prev->next, end);
        }
        return dummy.next;
    }
    ListNode *recur(ListNode *prev, ListNode *begin, ListNode *end) {
        ListNode *end_next = end->next;
        for (ListNode *p = begin, *cur = p->next, *next = cur->next; cur != end_next; p = cur, cur = next, next = next?next->next:NULL) {
            cur->next = p;
        }
        begin->next = end_next;
        prev->next = end;
        return begin;
    }
};