206. Reverse Linked List && 92. Reverse Linked List II

Reverse a singly linked list.
按顺序改变链表节点的指向，从指向下一个节点改为指向上一个节点。
即 本来 cur.next = next→cur.next = pre

 public ListNode reverseList(ListNode head) {
        if(head == null || head.next == null) return head;
        ListNode pre = null,cur = head,next = cur.next;
        while(true){
            cur.next = pre;
            pre = cur;
            cur = next;
            if(next == null) break;
            next = next.next;
        }
        return pre;
    }

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.

在上一题的基础上，只需要reverse部分链表。那就需要找到需要reverse的部分链表，reverse 的方法和上一题相同，然后将链表重新拼接在一起。

 public ListNode reverseBetween(ListNode head, int m, int n) {
      if(head == null || head.next == null || m == n) return head;
      // p、q指向需要reverse的部分链表的头、尾
      // start保存不需要reverse的前半部分的尾，end保存不需要reverse的后半部分的头，用于重新拼接成新的链表
        ListNode fakehead = new ListNode(0),p = fakehead,q = fakehead,start = head ,end = head;
        fakehead.next = head;
        for(int i = 0;i < m;i++){      
            start = p;
            p = p.next;
        }
         for(int i = 0;i < n;i++){
            q = q.next;
        }
        end = q.next;
       ListNode next = p.next.next,cur = p.next ,rp = p;  //reverse链表
        while(p != q){
            cur.next = p;
            p = cur;
            cur = next;
            if (next == null) {
                break;
            }
            next = next.next;
        }

        start.next = q;    //拼接
        rp.next = end;
        return fakehead.next;
    }