leetcode Rotate List

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1. 题目描述

Given a list, rotate the list to the right by k places, where k is non-negative.

For example:
Given 1->2->3->4->5->NULL and k = 2,
return 4->5->1->2->3->NULL.

2.解决方案1

class Solution {
public:
  ListNode *rotateRight(ListNode *head, int k)
{
  if(head == NULL || k <= 0)
    return head;
  //get length
  ListNode *temp = head;
  int node_count = 0;
  while(temp != NULL)
  {
    node_count++;
    temp = temp->next;
  }
  
  if(k > node_count)
     k = k%node_count;
  if(k == node_count || k == 0)
     return head;
  
  //first node move k step
  ListNode *first = head;
  while( k > 0)
  {
    first = first->next;
    k--;
  }
  
  //when first node move to last, the second node move (length - k) step
  ListNode *second = head;
  while(first->next != NULL)
  {
    first = first->next;
    second = second->next;
  }

  ListNode *newhead = second->next;
  //be a circle
 first->next = head;
  //broken circle
 second->next = NULL;
  return newhead;
}

};

3.解决方案2

class Solution {
public:
ListNode* getLastNode(ListNode *head, int& len){
	ListNode* node = head;
	len = 1;
	if(head == NULL){
	    len = 0;
	    return NULL;
	}
	while( node->next != NULL){
		node = node->next;
		len++;
	}
	return node;
}


    ListNode *rotateRight(ListNode *head, int k) {
       if(head == NULL){
		return head;
	}
	int len;
	ListNode* lastNode = getLastNode(head, len);
	lastNode->next = head;

	k %= len;
	int step = len - k;
	while(step>0){
		lastNode = lastNode->next;
		step--;
	}
	head = lastNode->next;
	lastNode->next = NULL;
	return head;
    }
};

leetcode Rotate List

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原文地址：http://blog.csdn.net/fox64194167/article/details/44086895