Reverse Nodes in k-Group

最新推荐文章于 2022-04-08 22:28:12 发布

Kehl

最新推荐文章于 2022-04-08 22:28:12 发布

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分类专栏： leetcode

本文链接：https://blog.youkuaiyun.com/Oliverkehl/article/details/39258531

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本文介绍了一种反转链表中节点的方法，每K个一组进行反转，并返回修改后的链表。讨论了当节点数不是K的倍数时，剩余节点保持原样的处理方式。文章通过示例展示了不同K值下的反转效果，并提供了使用虚拟头结点来简化代码的实现思路。

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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

练习使用虚拟头结点来简化code

/**
 * 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==NULL || k==1)
            return head;
        ListNode* node = new ListNode(-1);
        node->next = head;
        head = node;
        ListNode* q = node,*t = node;
        while(1){
            t = q;
            for(int i=0;i<k;i++){
                if(t==NULL)
                    return head->next;
                t = t->next;
            }
            if(t==NULL)
                return head->next;
            ListNode* res = t->next;
            t->next=NULL;
            q->next = reverse(q->next);
            for(int i=0;i<k;i++)
                q = q->next;
            q->next =res;
        }
    }
    ListNode* reverse(ListNode* head){
        if(head==NULL)
            return NULL;
        ListNode* t=NULL;
        while(head!=NULL){
            ListNode* node = head;
            head = head->next;
            node->next = t;
            t = node;
        }
        return t;
    }
};