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

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#### 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?

``` 1 class Solution {
2 public:
5         ListNode* pre = nullptr;
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);
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         }