LeetCode 25 Reverse Nodes in k-Group (Reverse node in K-Group Linked List)

Source: Internet
Author: User

LeetCode 25 Reverse Nodes in k-Group (Reverse node in K-Group Linked List)
Original

Given a linked list, the node of the linked list is reversed within a certain period of time, and the modified linked list is returned. If the number of knots is not a multiple of K, the remaining nodes remain unchanged. You should not modify the value of a node. Only the node itself can be modified. Only constant space is allowed. For example, for a given 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
Translation
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->5For k = 2, you should return: 2->1->4->3->5For k = 3, you should return: 3->2->1->4->5
Code

The following code is not mine. Although I think it is almost the same, it is almost the result difference ......

A long way to go!

/** * 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) {        auto node = head;        for(int i = 0; i < k; ++ i) {            if(!node) return head;            node = node->next;        }        auto new_head = reverse(head, node);        head->next = reverseKGroup(node, k);        return new_head;            }    ListNode* reverse(ListNode* start, ListNode* end) {        ListNode* head = end;        while(start != end) {            auto temp = start->next;            start->next = head;            head = start;            start = temp;        }        return head;    }};

 

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