19.1.30 [LeetCode 25] 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.

k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.

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

Note:

• Only constant extra memory is allowed.
• You may not alter the values in the list's nodes, only nodes itself may be changed.

题意

一个链表中每k个结点翻转，结尾处不满k个的保持原样

题解

class Solution {
public:
    ListNode* reverseKGroup(ListNode* head, int k) {
        if (k == 1)return head;
        int count = 0;
        ListNode*hh = NULL, *before = NULL, *p = NULL, *now = NULL;
        ListNode*end = head;
        for (int i = 0; i < k-1; i++)
            if(end)
                end = end->next;
        while (head) {
            count++;
            if (count == 1 && !end)
            {
                if (!hh)
                    hh = head;
                break;
            }
            if (count == k) {
                if(before)
                    before->next = head;
                before = now;
            }
            if (count == 1) 
                now = head;
            if (hh == NULL && count == k)hh = head;
            ListNode*tmp = head->next;
            head->next = p;
            p = head;
            head = tmp;
            if (count == k) {
                count = 0;
                p = NULL;
            }
            if(end)
                end = end->next;
        }
        if(before)
            before->next = head;
        return hh;
    }
};

我的题解考虑了很多特殊情况，估计有简单得多的解法吧，现在乱七八糟的