25. Reverse Nodes in k-Group（K 个一组，反转链表）

25. Reverse Nodes in k-Group


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.

You may not alter the values in the list's nodes, only nodes themselves may be changed.

 

Example 1:

Input: head = [1,2,3,4,5], k = 2
Output: [2,1,4,3,5]

Example 2:

Input: head = [1,2,3,4,5], k = 3
Output: [3,2,1,4,5]

Example 3:

Input: head = [1,2,3,4,5], k = 1
Output: [1,2,3,4,5]

Example 4:

Input: head = [1], k = 1
Output: [1]

 

Constraints:

  • The number of nodes in the list is in the range sz.
  • 1 <= sz <= 5000
  • 0 <= Node.val <= 1000
  • 1 <= k <= sz

 

Follow-up: Can you solve the problem in O(1) extra memory space?
 

 

 

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reverseKGroup(self, head: Optional[ListNode], k: int) -> Optional[ListNode]:

        def reverse(head,end):
            pre = None
            cur = head
            while cur != end:
                nxt = cur.next
                cur.next = pre
                pre = cur
                cur = nxt
            return pre

        if head == None:
            return
        end = head
        for i in range(k):
            if end == None:
                return head
            end = end.next
        new = reverse(head,end)
        head.next = self.reverseKGroup(end,k)
        return new

 

 1 /**
 2  * Definition for singly-linked list.
 3  * struct ListNode {
 4  *     int val;
 5  *     ListNode *next;
 6  *     ListNode() : val(0), next(nullptr) {}
 7  *     ListNode(int x) : val(x), next(nullptr) {}
 8  *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 9  * };
10  */
11 class Solution {
12 public:
13     ListNode* reverseList(ListNode* head,ListNode* end) {
14         //[head,end)
15         if(head == nullptr || head->next == nullptr) return head;
16         ListNode* pre = nullptr;
17         ListNode* cur = head;
18         while(cur != end) {
19             ListNode* c_next = cur->next;
20             cur->next = pre;
21             pre = cur;
22             cur = c_next;
23         }
24         return pre;
25     }
26     ListNode* reverseKGroup(ListNode* head, int k) {
27         if (head == nullptr) return nullptr;
28         ListNode* end = head;
29         for(int i = 1; i <= k;i++) {
30             if (end == nullptr) return head;
31             end = end->next;
32         }
33         ListNode* newhead = reverseList(head,end);
34         head->next = reverseKGroup(end,k);
35         return newhead;
36     }
37 };

 

 
 
 
 
 
 
 
 
 
 

 

 

 
 
 1 class Solution {
 2 public:
 3     ListNode* reverseList(ListNode* head) {
 4         if(head == nullptr || head->next == nullptr) return head;
 5         ListNode* pre = nullptr;
 6         ListNode* cur = head;
 7         while(cur != nullptr) {
 8             ListNode* c_next = cur->next;
 9             cur->next = pre;
10             pre = cur;
11             cur = c_next;
12         }
13         return pre;
14     }
15     ListNode* reverseKGroup(ListNode* head, int k) {
16         ListNode* fakehead = new ListNode(-1);
17         fakehead->next = head;
18         ListNode* pre =fakehead;
19         ListNode* cur = head;
20         while(cur != nullptr) {
21             for(int i = 1;i < k && cur != nullptr ;++i) {
22                 cur = cur->next;
23             }
24             if(cur==nullptr) break;
25             
26             ListNode* start = pre->next;
27             ListNode* c_next = cur->next;
28 
29             cur->next = nullptr;
30             pre->next =  reverseList(start);
31             start->next = c_next;
32             pre = start;
33             cur = pre->next;
34         }
35         return  fakehead->next;
36     }
37 };

 

posted @ 2021-09-12 12:31  乐乐章  阅读(44)  评论(0)  编辑  收藏  举报