Reverse Nodes in k-Group

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.

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：用一个栈保存需要逆序的k个节点，然后利用栈FILO原则，就可以每K个倒置一次
 */
/**
 * 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 *>Q;
	    ListNode *ans=new ListNode(0);
		ans->next=head;
		ListNode *answer=ans;
		while(ans && ans->next){
		ListNode *cur=ans;
		int n=k;
		  while(n--){
	    	   if(cur->next==NULL){
				   break;
			   }
			  Q.push(cur->next);
			  cur=cur->next;
		  }
		  if(Q.size()<k){
			  break;
		  }
		  ListNode *ans_next=Q.top()->next;
		  while(Q.size()){
		  ans->next=Q.top();
		  Q.pop();
		  ans=ans->next;
		  }
		  ans->next=ans_next;
		}
     return answer->next;   
 }
};