[LeetCode 解題報告]025. 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.

考察：與206. Reverse Linked List相似，在逆置單鏈表的基礎上，每k個進行逆置，所以首先需要統計單鏈表的長度，然後沒k個進行逆置，小於k個則不逆置；

/**
 * 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) {
        ListNode *dummy = new ListNode(-1);
        dummy->next = NULL;
        
        ListNode *cur = head;
        int nums = 0;
        while(cur) {
            nums += 1;
            cur = cur->next;
        }
        
        ListNode *pre = dummy, *last = dummy;
        cur = head;
        while(nums >= k) {
            last = cur;
            for (int i = 0; i < k; i ++) {
                ListNode *t = cur;
                cur = cur->next;
                t->next = pre->next;
                pre->next = t;
            }
            pre = last;
            nums -= k;
        }
        pre->next = cur;
        return dummy->next;
    }
};

完，