[LeetCode练习题-C语言]25. Reverse Nodes in k-Group

题目

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

我的答案

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     struct ListNode *next; * }; */struct ListNode* reverseKGroup(struct ListNode* head, int k) {    struct ListNode* node = head;    struct ListNode* temp;    int i;    int *valList;    while(node != NULL){        temp = node;        for(i=0; i<k; i++){            if(node != NULL){                valList[i] = node->val;                node = node->next;            }else{                k=0;                break;            }        }        for(i=0; i<k; i++){            temp->val = valList[k-i-1];            temp = temp->next;        }    }    return head;}

思路

五个节点为一组，将值保存进数组中，然后进行交换。

别人的代码

使用了递归

/** * 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;    }};