第六天leetcode刷题
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.
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.
具体代码实现如下:
/**
* 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)
return head;
ListNode* current = head,*next1,*pre = nullptr;
int m = 1;
while(m <= k && current)
{
next1 = current ->next;
current ->next = pre;
pre = current;
current = next1;
++m;
}
if(m <= k)
{
while(current != head)
{
ListNode* temp = pre ->next;
pre ->next = current;
current = pre;
pre = temp;
}
pre = head;
}
else
head ->next = reverseKGroup(current,k);
return pre;
}
};