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
/**
* 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) return NULL;
ListNode *sentry = new ListNode(0);
sentry->next = head;
ListNode *p, *q, *r;
p = sentry;
while (p){
int i = 0;
q = p;
while (q->next && i <k){
q = q->next;
i++;
}
if (i != k) return sentry->next;
i = 1;
r = p->next;
while (i<k){
ListNode* tmp;
tmp = r->next;
r->next = r->next->next;
tmp->next = p->next;
p->next = tmp;
i++;
}
p = r;
}
return sentry->next;
}
};

本文介绍了一种算法,该算法可以将链表中的节点每K个一组进行反转,并返回修改后的链表。如果节点数量不是K的倍数,剩余节点保持原样。此操作不改变节点值,仅调整节点顺序,且只使用常量内存。
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