0025._Reverse_Nodes_In_K_Group.md

25.reverse nodes in k group

难度Hard

刷题内容

原题连接

• https://leetcode.com/problems/reverse-nodes-in-k-group/

内容描述

Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.

k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.

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

Note:

Only constant extra memory is allowed.
You may not alter the values in the list's nodes, only nodes itself may be changed.

思路1 - 时间复杂度: O(N)- 空间复杂度: O(1)******

题目本身不难，这题我用了递归的方法去解，但题目中的note中说不能有额外的储存空间，但递归会生成辅助的空间，可以把递归改成循环，不过接下还是递归的版本

/**
 * 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)
        return head;
    ListNode* current = head,*next1,*pre = nullptr;
    int m = 1;
    while(m <= k && current)
    {
        next1 = current ->next;
        current ->next = pre;
        pre = current;
        current = next1;
        ++m;
    }
    if(m <= k)
    {
        while(current != head)
        {
            ListNode* temp = pre ->next;
            pre ->next = current;
            current = pre;
            pre = temp;
        }
        pre = head;
    }
    else
        head ->next = reverseKGroup(current,k);
    return pre;
    }
};