第20题 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.

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 reverseBetween(ListNode head, int m, int n) {
        if(m==n) return head;
        ListNode fakeHead = new ListNode(0);
        fakeHead.next=head;
        ListNode first = fakeHead;
        int i=0;
        for(; i<m-1; i++){
            first=first.next;
        }
        ListNode tail=first.next;
        ListNode prev = tail, cur=prev.next, next;
        for(i=m;i<n;i++) {
            next=cur.next;
            cur.next=prev;
            prev=cur;
            cur=next;        
        }
        
        first.next=prev;
        tail.next=cur;
        return fakeHead.next;
    }
}

Note：先保存下点m-1。从点m到点n，将list反序，在这过程中需要保存反序部分的尾部tail，反序最后一步的prev节点就是反序节点的头部。然后将反序的list接入原list中，between节点m-1和n+1。