LeetCode 61 Rotating Linked List (C++)

Table of contents

Title description:

Problem-solving ideas:

Method: fast and slow pointer

Illustration:

Specific operation:

 Specific code:

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Title description:

Problem-solving ideas:

Method: fast and slow pointer

Illustration:

Example: head=[1,2,3,4,5],k=2;

step:

Specific operation:

• First, we judge whether the linked list is empty, if it is empty, return head, if it is not empty, perform the following operations;
• Define a fast pointer fast and a slow pointer slow;

  ListNode* fast=head;
  ListNode* slow=head;

• Let the fast pointer (fast) go k steps first;

  for(int i=0;i>k;i++){
      fast=fast->next;
  }

• At this point we have to consider whether k is greater than the length of the linked list , if it is less than the length of the linked list: if k is less than the length of the linked list, it is k in the for loop in the previous step; if k is greater than the length of the linked list, k in the for loop in the previous step must be calculated.
• According to the law, we found that k=k% the length of the linked list can be made , so that the two conditions are satisfied. We use a new function to calculate the length of the array;

  int fun(LinsNode* head){
      int len=0;
      while(head){
          head=head->next;
          len++;
      }
      return len;
  }

• Then we make k=k%len execute the for loop;
• At this time, there is a difference of k bits between the fast (fast) and slow (slow) pointers, and then let them move together, and end the loop when fast->next == nullptr;

  while(fast->next){
      fast=fast->next;
      slow=slow->next;
  }

• The fast pointer (fast) goes to the end of the linked list, and then the next node points to the head node; the new head node points to slow; slow refers to empty and becomes the tail

  fast->next=head; 
  head=slow->next; 
  slow->next=-nullptr;

 Specific code:

class Solution {
public:
    //先考虑k是否大于链表长度
    int fun(ListNode* head){
        int len=0;
        while(head){
            head=head->next;
            len++;
        }
        return len;
    }
    ListNode* rotateRight(ListNode* head, int k) {
        if(head==nullptr)return head;
        //计算节点数
        int len=fun(head);
        k = k % len;
        //快慢指针
        ListNode* fast=head;
        ListNode* slow=head;
        //快指针先走k步
        for(int i=0;i<k;i++){
            fast=fast->next;
        }
        //快慢指针同时移动，直到fast->next==nullptr
        while(fast->next){
            fast=fast->next;
            slow=slow->next;
        }
        //快指针到链表尾部，然后下一个结点指向头结点
        fast->next=head;
        //新的头结点指向slow
        head=slow->next;
        //slow指针指空，变成尾部
        slow->next=nullptr;
        return head;
    }
};