标签：alt tno link length == code note const tip
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:
code:
using a stack
/**
* 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) {
stack<ListNode*> s;
ListNode* dummy = new ListNode(0);
dummy->next = head;
ListNode* current = dummy;
ListNode* next = dummy->next;
while (next) {
for (int i = 0; i < k && next != NULL; ++i) {
s.push(next);
next = next->next;
}
if (s.size() != k) return dummy->next;
while (!s.empty()) {
current->next = s.top();
s.pop();
current = current->next;
}
current->next = next;
}
return dummy->next;
}
};
Runtime: 24 ms, faster than 32.59% of C++ online submissions for Reverse Nodes in k-Group.
this isn‘t conform with this question‘s demand.
the right way:
class Solution {
public:
ListNode *reverseKGroup(ListNode *head, int k) {
if(head==NULL||k==1) return head;
int num=0;
ListNode *preheader = new ListNode(-1);
preheader->next = head;
ListNode *cur = preheader, *nex, *pre = preheader;
while(cur = cur->next)
num++;
while(num>=k) {
cur = pre->next;
nex = cur->next;
for(int i=1;i<k;++i) {
cur->next=nex->next;
nex->next=pre->next;
pre->next=nex;
nex=cur->next;
}
pre = cur;
num-=k;
}
return preheader->next;
}
};
Runtime: 28 ms, faster than 16.72% of C++ online submissions for Reverse Nodes in k-Group.
it will have a difficulty to understand.
标签：alt tno link length == code note const tip
原文地址：https://www.cnblogs.com/ruruozhenhao/p/9751893.html
