Reverse Nodes in K group

LeetCode #25

Description:

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.

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

Only constant memory is allowed.

Example:

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

Idea:

k个group一起reverse，reverse的时候用到三个指针，前一个，头，尾。 reverseList(ListNode *before, ListNode* begin, ListNode* end)

Code:

class Solution {
public:
    ListNode* reverseKGroup(ListNode* head, int k) {
        if(head==NULL || head->next==NULL || k<2) return head;
        ListNode dummy(-1);
        ListNode *pre=&dummy;
        pre->next=head;
        ListNode *end=head;
        while(end!=NULL){
            for(int i=1; i<k && end!=NULL; ++i){
                end=end->next;
            }
            if(end==NULL) break;
            pre=reverseList(pre, pre->next, end);
            end=pre->next;
        }
        return dummy.next;
    }

    ListNode* reverseList(ListNode *before, ListNode* begin, ListNode* end) {
        ListNode *end_next=end->next; // 千万要存这个！不要直接用end->next!
        ListNode *pre=end_next, *curr=begin;
        ListNode *tmp;

        while(curr!=end_next){
            tmp = curr->next;
            curr->next=pre;
            pre=curr;
            curr=tmp;
        }
        before->next=pre;
        return begin;

    }
};