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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
每次取k个长度的链表进行反转。时间：30ms
代码如下：
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: bool isKLen(ListNode* head, int k){ if (k <= 0 || head == NULL) return false; ListNode *phead = head; int count = k - 1; while (phead->next != NULL&&count){ phead = phead->next; count--; } if (count == 0) return true; else return false; } ListNode* reverseKGroup(ListNode* head, int k) { if (head == NULL || k <= 1) return head; ListNode *p, *q, *phead, *pnext = new ListNode(0); pnext->next = p = head; phead = pnext; while (isKLen(p,k)){ int count = k-1; ListNode *qt = pnext->next; while (count--){ q = p->next; p->next = q->next; q->next = qt; qt = q; } pnext->next = qt; pnext = p; p = p->next; } return phead->next; } };
[LeetCode] #25 Reverse Nodes in k-Group
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原文地址：http://www.cnblogs.com/Scorpio989/p/4549794.html
