25. Reverse Nodes in k-Group

Posted 2021-01-11 ruruozhenhao

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.

 

code:

using a stack

/**
 * 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) {
        stack<ListNode*> s;
        ListNode* dummy = new ListNode(0);
        dummy->next = head;
        ListNode* current = dummy;
        ListNode* next = dummy->next;
        while (next) {
            for (int i = 0; i < k && next != NULL; ++i) {
                s.push(next);
                next = next->next;
            }
            if (s.size() != k) return dummy->next;
            while (!s.empty()) {
                current->next = s.top();
                s.pop();
                current = current->next;
            }
            current->next = next;
        }
        return dummy->next;
    }
};

Runtime: 24 ms, faster than 32.59% of C++ online submissions for Reverse Nodes in k-Group.

this isn‘t conform with this question‘s  demand.

 

the right way:

class Solution {
public:
    ListNode *reverseKGroup(ListNode *head, int k) {
        if(head==NULL||k==1) return head;
        int num=0;
        ListNode *preheader = new ListNode(-1);
        preheader->next = head;
        ListNode *cur = preheader, *nex, *pre = preheader;
        while(cur = cur->next) 
            num++;
        while(num>=k) {
            cur = pre->next;
            nex = cur->next;
            for(int i=1;i<k;++i) {
                cur->next=nex->next;
                nex->next=pre->next;
                pre->next=nex;
                nex=cur->next;
            }
            pre = cur;
            num-=k;
        }
        return preheader->next;
    }
};

Runtime: 28 ms, faster than 16.72% of C++ online submissions for Reverse Nodes in k-Group.

it will have a difficulty to understand.