22nd question rotate list

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

For example:
Given1->2->3->4->5->NULLAnd k =2,

Return4->5->1->2->3->NULL.

Solution:

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { *         val = x; *         next = null; *     } * } */public class Solution {    public ListNode rotateRight(ListNode head, int n) {        if(head==null||head.next==null || n==0) return head;        ListNode fakeHead = new ListNode(0);        fakeHead.next=head;        int length=0;        ListNode start=fakeHead, end=fakeHead;        for(int i=0; i<n; i++){            if(end.next==null) break;            end=end.next;            length++;        }        if(end.next==null){            //when n>=length            //int startPos = (length-n)%length+length;      //Or:            int startPos=n%length;            if(startPos==0) return fakeHead.next;            startPos=length-startPos;                        for(int i=0; i<startPos; i++){                start=start.next;            }        }        else{            while(end.next!=null){                start=start.next;                end=end.next;            }        }        //reorder list          end.next=fakeHead.next;        fakeHead.next=start.next;        start.next=null;        return fakeHead.next;                    }}

NOTE: If end. Next is not null after the end is shifted n times, the length of the list is greater than N. Then start and end are moved back.

Otherwise, the length of the list is <= n, and the actual number of rotate times must be found at this time. For a long list, if the number of rotate is length, the result is the original list. Therefore, the actual number of Rotate operations is n % length. If the result is 0, the original list is directly returned. Otherwise, the corresponding start point must be found, that is, the starting point of the list change.

Note:

 int startPos = (length-n)%length+length;      

Actually and

int startPos=n%length;startPos=length-startPos;

One effect (when startpos is not 0 ). However, the latter is easier to understand than the hacker.

The modulo of another negative number to a positive number is calculated as follows:

Result = m % n, where M <= 0, N> 0

Assume M = xn + result, where x <= 0, result <0 and-result <n.

For example (-5) % 3. -5 = 3 * (-1) + (-2 ). So (-5) % 3 =-2.

22nd question rotate list