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;
}

};


思路：使用了快慢指针的技巧。但由于k的值有可能大于整个的长度，所以还要先求出list的长度。先用快的指针走k步，然后再把它走到尾部，这个时候慢指针就走了length - k 的步数。再头尾相连组成环，再断环，返回新的头部指针。


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;
    }
};


思路：先直接一个一个走到最后一个指针，然后连接成环，然后再行走lenght - k 的步数。再断环。