Leetcode: 25. Reverse Nodes in k-Group

Description

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 nodes, only nodes itself may be changed.

Only constant memory is allowed.

Example

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

思路

• k-group反转
• 首先写一个单链表反转函数，指定头结点和尾节点，然后再函数中反转链表
• 从给定链表头开始，每次读取k个长度的链表，调用反转函数，然后再继续这个过程
• 注意考虑两个反转链表之间的连接时候的情况

代码

/**
 * 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) {
        if(k == 1) return head;
        
        ListNode *res = NULL, *ptr = NULL, *cur = NULL, *pre = NULL;
        while(head){
            int count = k;
            ptr = NULL;
            cur = head;
            //while实现每次截取k个长度的链表
            while(head){
                if(ptr == NULL)
                    ptr = head;
                    
                cur = head;
                head = head->next;
                count--;
                if(count == 0) break;
            }
            
            //此时需要反转
            if(count == 0 && cur && ptr){
                Reverse(&ptr, &cur);
            } 
            
            //连接两个反转部分或不反转部分
            if(!pre){
                pre = cur;
            }
            else{
                pre->next = ptr;
                pre = cur;
            }
         
            //记录反转链表的头节点
            if(!res) res = ptr;
            
            //特殊情况考虑，即刚好反转到链表末尾
            if(!head && count == 0) pre->next = NULL;
        }
        
        return res;
    }
    
    //反转单链表
    void Reverse(ListNode **head, ListNode **tail){
        ListNode *end = *tail, *cur = *head, *ptr = NULL, *tmp = NULL;
       
        ptr = cur->next; 
        *tail = cur;
        while(ptr != end){
            tmp = ptr->next;
            ptr->next = cur;
            cur = ptr;
            ptr = tmp;
        }
        
        if(ptr != end)
            end->next = ptr;
        else
            end->next = cur;
           
        *head = end;
    }
};