每天一道算法题

反转链表，leetcode 206题。

思路：
定义三个pre、current、next三个节点。每次遍历需要做的事：
1.next节点赋值(current.next)
2.反转当前节点指针(current.next = pre)
3.pre节点和current节点步进到下一节点

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode reverseList(ListNode head) {
        ListNode pre = null;
        ListNode current = head;
        ListNode next;
        while(current != null) {
            next = current.next;
            current.next = pre;

            pre = current;
            current = next;
        }
        return pre;
    }
}