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
思路： 先实现reverseList（head）， 然后实现reverse（start， end）反正从start到end的链表，然后基于reverse（start，end）来完成最后的动作。
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode *reverse(ListNode* head) { if(head == NULL) return NULL; ListNode * pre = NULL; ListNode * cur = head; ListNode * next = NULL; // make cur->next = pre; while(cur) { next = cur->next; cur->next = pre; pre = cur; cur = next; } return pre; } #if 1 ListNode* reverse(ListNode* head, ListNode* end) { if(head == NULL) return NULL; ListNode * pre = NULL; ListNode * cur = head; ListNode * next = NULL; // make cur->next = pre; while(cur) { next = cur->next; cur->next = pre; pre = cur; cur = next; if(pre == end) break; } return pre; } ListNode *reverseKGroup(ListNode *head, int k) { if(head == NULL) return NULL; ListNode dummy(-1); dummy.next = head; ListNode* pre = &dummy; ListNode* start= head; ListNode* end = start; ListNode* next= NULL; int cnt = 0; while(end) { cnt++; if(cnt == k) { next = end->next; //assign pre->next = reverse(start, end); start->next = next; //cout << "start\t" << start->val <<endl; //cout << "end\t" << end->val <<endl; //printList(dummy.next); //printList(pre->next); //update pre = start; start = next; end = next; cnt = 0; //cout << "start\t" << start->val <<endl; //cout << "end\t" << end->val <<endl; //cout << "pre\t" << pre->val <<endl; } else end = end->next; } return dummy.next; } #endif };
[LeetCode] Reverse Nodes in k-Group
原文：http://www.cnblogs.com/diegodu/p/4280031.html