LeetCode 25: Reverse Nodes in k-Group

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本文介绍了一种算法，该算法接收一个链表及整数K作为输入，然后将链表中的节点每K个一组进行翻转。如果剩余节点不足K个，则保持不变。文章详细解释了如何通过迭代实现这一功能，特别强调了只允许使用常量级额外内存的限制条件。

Reverse Nodes in k-Group

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

解题思路

/**
 * 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 (k < 2) return head;

        // 设置一个额外的表头指针
        ListNode result(0), *tail = &result;

        int count = 0;
        ListNode *begin = head;
        while (head) {
            count++;

            if (count == k) {
                // 将 begin 和 head 之间的子链表反转，并链到 tail 结点后边，更新 tail 指针
                ListNode *p = begin, *q = begin->next, *end = head->next;
                while (q != end) {
                    ListNode *temp = q->next;
                    q->next = p;
                    p = q;
                    q = temp;
                }

                tail->next = p;
                tail = begin;

                count = 0;
                head = begin = q;
            }
            else {
                head = head->next;
            }
        }

        tail->next = begin;
        return result.next;
    }
};