25. 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.

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

解法：

采用递归法或者迭代法

/**
* 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) {
if(!head || !head->next || k < 2) return head;
ListNode* next_group = head;
for(int i =0 ;i < k ; ++i){
if(next_group) next_group =next_group->next;
else return head;
}
ListNode* next_head = reverseKGroup(next_group,k);
ListNode *pre = NULL , *cur = head;
while(cur != next_group) {
ListNode* next = cur->next;
cur->next = pre?pre:next_head;
pre = cur;
cur = next;
}
return pre;
}
};

class Solution {
public:
istNode *reverseKGroup(ListNode *head, int k) {
if (head == nullptr || head->next == nullptr || k < 2) return head;
ListNode dummy(-1);
dummy.next = head;
for(ListNode *prev = &dummy, *end = head; end; end = prev->next) {
for (int i = 1; i < k && end; i++)
end = end->next;
if (end == nullptr) break; // 不足 k 个
prev = reverse(prev, prev->next, end);
}
return dummy.next;
}
// prev 是 first 前一个元素, [begin, end] 闭区间，保证三者都不为 null
// 返回反转后的倒数第 1 个元素
ListNode* reverse(ListNode *prev, ListNode *begin, ListNode *end) {
ListNode *end_next = end->next;
for (ListNode *p = begin, *cur = p->next, *next = cur->next;
cur != end_next;p = cur, cur = next, next = next ? next->next : nullptr) {
cur->next = p;
}
begin->next = end_next;
prev->next = end;
return begin;
}
};