[leedcode 92] Reverse Linked List II

Reverse a linked list from position m to n. Do it in-place and in one-pass.

For example:
Given 1->2->3->4->5->NULL, m = 2 and n = 4,

return 1->4->3->2->5->NULL.

Note:
Given m, n satisfy the following condition:
1 ≤ m ≤ n ≤ length of list.

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode reverseBetween(ListNode head, int m, int n) {
        /*The idea is stright forward. There needs an safe head guard.

        1 calculate the number that needs to reverse 

        2 find the start position to reverse.

        3. iterative reverse two node(Use two pointer)*/
        ListNode newHead=new ListNode(-1);//安全头节点
        newHead.next=head;
        ListNode p1=head;
        ListNode last=newHead;
        int t=n-m;
        while(m>1){
            p1=p1.next;
            last=last.next;
            m--;
        }
        ListNode rear=last;
        ListNode temp=p1;
            p1=p1.next;
            last=last.next;
        ListNode p2=p1;
        
        while(t>0){
            p2=p1.next;
            p1.next=last;
            last=p1;
            p1=p2;
            t--;
        }
        rear.next=last;
        temp.next=p1;
        return newHead.next;
    }
}