leetcode 25. Reverse Nodes in k-Group

最新推荐文章于 2020-12-29 09:44:23 发布

DrawnBreak

最新推荐文章于 2020-12-29 09:44:23 发布

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本文链接：https://blog.youkuaiyun.com/a799581229/article/details/52442889

本文介绍了一种算法，该算法可以实现链表中每K个节点进行一次翻转的操作，并保持剩余节点顺序不变。文章详细展示了如何通过迭代方式完成这一任务，包括设置临时指针、调整节点指向等关键步骤。

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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

这题做的我有点炸，，花了一个多小时才写对，，脑子乱掉了。。就不写思路了

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* reverseKGroup(struct ListNode* head, int k) {
    struct ListNode *pleft,*pmid,*pright,*plast,*pnew,*new_head;
    int i;
    if(head==NULL||head->next==NULL||k==1)
        return head;
    
    pnew=head;    
    for(i=1;i<=k-1;i++)  //寻找第k个结点，并设为头节点
    {
        pnew=pnew->next;
        if(pnew==NULL)
            return head;
    }
    new_head=pnew; //第k个为头节点
    
    
    pleft=head;  //left
    pmid=head->next;  //mid
    plast=head;
        
    while(pmid!=NULL)
    {
        pright=pmid->next;  //pright
        pmid->next=pleft;   //mid指向left
        
        if(pmid==pnew)   //如果mid是第k个节点
        {
            pmid=pright;   //pmid变成pright
            pnew=pmid;     //pnew从当前开始算
            plast->next=pnew;  //在判断是否还有k个时，先将plast指向当前这个
            if(pnew==NULL)   //如果后面已经全为空了，那么可以直接输出了
                return new_head;
            for(i=1;i<=k-1;i++)   //否则就继续寻找第k个
            {
                pnew=pnew->next;
                if(pnew==NULL)  //如果出现空，则直接返回，不用逆序
                    return new_head;
            }
            //若有k个，则plast指向第k个
            plast->next=pnew;  //plast一定是指向它
            plast=pmid;  //重置plast
            continue;
        }
        
        pleft=pmid;
        pmid=pright;
    }
    return new_head;
}