LEETCODE ALGORITHM:25. Reverse Nodes in k-Group

题目

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.

题解

添加头结点，避免单独处理第一组数据

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
private:
    pair<ListNode*,ListNode*> reve(ListNode* head,ListNode* tail)
    {
        auto pre=tail->next;
        auto p=head;
        while(pre!=tail)
        {
            auto next=p->next;
            p->next=pre;
            pre=p;
            p=next;
        }
        return{tail,head};
    }
public:
    ListNode* reverseKGroup(ListNode* head, int k) {
        ListNode* h=new ListNode(0);
        h->next=head;
        auto pre=h;
        while(h)
        {
            auto tail=pre;
            for(int i=0;i<k;++i)
            {
                tail=tail->next;
                if(!tail) return h->next;
            }
            auto ne=tail->next;

            tie(head,tail)=reve(head,tail);
            pre->next=head;
            tail->next=ne;

            head=tail->next;
            pre=tail;
        }
        return h->next;
    }
};