leetcode:Reverse Nodes in k-Group

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本文介绍了一种算法，用于在保持节点值不变的情况下，批量反转链表中连续K个节点。通过此方法，可以有效地操作链表结构，特别适用于需要对节点进行周期性操作的场景。

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
分析
从头节点到尾节点，依次翻转K个节点，翻转K个节点时，记录K个翻转后节点的头节点指针和尾节点指针。
C++代码

class Solution {
public:
    ListNode* reverseKGroup(ListNode* head, int k) {
        ListNode*p,*q1=head,*lastNode=head,*q2,*last;
        if (!head||k==0)
            return head;
        q2 = nodeAfterKNodes(q1, k);
        head = reverseList(q1, k, last);
        lastNode = last;
        q1 = q2;
        while (q1){ 
            q2 = nodeAfterKNodes(q1, k);
            lastNode->next= reverseList(q1, k, last);
            lastNode = last;
            q1 = q2;
        }
        return head;
    }
    ListNode* reverseList(ListNode* head,int k,ListNode*& last){
        if (!head || k > length(head)){
            last = head;
            return head;
        }   
        ListNode* p = head, *q=NULL, *r=NULL;
        last = head;
        p = head;
        q = head->next;
        head->next = NULL;
        if (q)
            r = q->next;
        int i = 0;
        while (q && i<k-1){
            q->next = p;
            p = q;
            q = r;
            if (r)
                r = r->next;
            i++;
        }
        return p;
    }
    int length(ListNode* head){
        int len = 0;
        while (head){
            len++;
            head = head->next;
        }
        return len;
    }
    ListNode* nodeAfterKNodes(ListNode* head, int k){
        if (k > length(head))
            return NULL;
        int i = 0;
        while (i < k){
            head = head->next;
            i++;
        }
        return head;
    }
};