LeetCode-25- Reverse Nodes in k-Group

算法描述：

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.

解题思路：细节实现题。

ListNode* reverseKGroup(ListNode* head, int k) {
    if(head == nullptr)  return head;
    int count =0;
    ListNode* cur = head;
    while(cur!=nullptr && count < k){
        cur=cur->next;
        count++;
    }
    if(count==k){
        cur = reverseKGroup(cur,k);
        while(count--){
            ListNode* temp = head->next;
            head->next=cur;
            cur=head;
            head=temp;
        }
        head=cur;
    }
    return head;
}

LeetCode-25- Reverse Nodes in k-Group

原文：https://www.cnblogs.com/nobodywang/p/10362104.html