LeetCode:Reverse Nodes in k-Group

Reverse Nodes in k-Group

   

Total Accepted: 57696  Total Submissions: 210241  Difficulty: Hard

Given a linked list, reverse the nodes of a linked list k at a time and return its modified 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.

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

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思路：

k=3

第一次反转，初始状态：
 0 -> 1 -> 2 -> 3 -> 4 -> 5
 |    |    | 
pre  cur  tmp

第1次操作后：
 0 -> 2 -> 1 -> 3 -> 4 -> 5
 |         |    | 
pre       cur  tmp
第2次操作后：
 0 -> 3 -> 2 -> 1 -> 4 -> 5
 |              |    | 
pre            cur  tmp

第二次反转，初始状态：
 0 -> 3 -> 2 -> 1 -> 4 -> 5
                |    |    | 
               pre  cur  tmp

...

code:

/**
 * 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(!head || k<=1) return head;
        
        ListNode dummy(0);
        dummy.next = head;
        ListNode *pre = &dummy;
        ListNode *cur = head;
        ListNode *tmp;
        
        int len = 0;
        while(cur){ // 计算链表长度
            len++;
            cur = cur->next;
        }
        
        while(len>=k){
            cur = pre->next;
            tmp = cur->next;
            for(int i=1;i<k;i++){
                cur->next = tmp->next;
                tmp->next = pre->next;
                pre->next = tmp;
                tmp = cur->next;
            }
            pre = cur;
            len -= k;
        }
        return dummy.next;
    }
};