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.

相关问题是

Reverse Nodes in k-Group

 

 

public class Solution {

    public ListNode reverseBetween(ListNode head, int m, int n) {

        ListNode h = new ListNode(-1);

        h.next = head;

        ListNode pre = h;

        int cnt = 1;

        while(cnt<m){

             pre = pre.next;

             cnt++;

        }

        ListNode mnode = pre.next;

        ListNode nnode = mnode;

        while(cnt<n){

            nnode = nnode.next;

            cnt++;

        }

        ListNode end = nnode.next;

        reverse(pre, end);

        return h.next;

    }

    

    public void reverse(ListNode pre, ListNode end){ // B keep to being inserted after pre  pre-1-2-3-end, 1= A, 2=B, 3=t

        ListNode A = pre.next, B  = A.next; // A 定义为B前面一位, 不断的把B塞到pre后面（也就是list的第一位，然后把原来B的前面A链接到B的后面t上去

        while( B!=end){

            ListNode t =  B.next;

            B.next = pre.next;

            pre.next = B;

            A.next = t;

            B=t;

        }

    }

}