Reverse Nodes in k-Group

最新推荐文章于 2022-04-08 22:28:12 发布

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最新推荐文章于 2022-04-08 22:28:12 发布

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本文链接：https://blog.youkuaiyun.com/zzyafyj/article/details/41685157

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本文介绍了一种算法，该算法将给定的链表每K个节点为一组进行翻转，并返回修改后的链表。如果节点总数不是K的倍数，则剩余节点保持不变。该方法不允许改变节点的值，仅允许改变节点之间的连接方式。

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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

/**
 * 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) {
        int len = length(head);
        ListNode *h=head;
        if(len>=k){
            stack<ListNode *> stk;
            for(int i= 1;i<=k;++i){//保存需要反转的结点
                stk.push(head);
                head=head->next;
            }
            ListNode *p=NULL;
            if(!stk.empty()){//做反转
                p=stk.top();
                h = p;
                stk.pop();
                while(!stk.empty()){
                    ListNode *q = stk.top();
                    stk.pop();
                    p->next = q;
                    p=q;
                }
            }
            if(p)
               p->next = reverseKGroup(head,k);
        }
        return h;
    }
private:
   int length(ListNode *head){
       int i=0;
       while(head){
           ++i;
           head=head->next;
       }
       return i;
   }
   
};