[LeetCode]Reverse Nodes in k-Group_reversenodesinkgroup-优快云博客

本文介绍了一种高级链表操作算法，详细解释了如何将链表中的节点按K组进行翻转，并提供了代码实现。适用于链表处理和数据结构优化。

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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

分析：翻转链表的高级版，注意翻转几次，采用四个指针动态移动。每次把链表最后的一个数放到最前面。

/**
 * 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 *cur = head;
        ListNode *next = head;
        ListNode *temp = head;
        ListNode *pre = new ListNode(-1);
        pre->next = head;
        ListNode *ret = pre;
        int n = 0;
        while(cur){
            ++n;
            cur = cur->next;
        }
        while(n>=k){
            cur=pre->next;
            next=cur->next;
            for(int i=1;i<k;++i){
                temp = next->next;
                next->next=pre->next;
                pre->next=next;
                cur->next=temp;
                next = temp;
            }
            pre = cur;
            n = n-k;
        }
        return ret->next;
    }
};