Reverse Linked List

题目地址：
https://leetcode.com/problems/reverse-linked-list/#/description

题目：
Reverse a singly linked list.

解题思路：
这道题主要是需要掌握两种方法：iterative和recursive。这道题还会有一个follow up：

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.

对于follow up，需要的是加上dummy，pre，start和then。首先找到要开始反转的起点的前一个node设为pre，然后把开始反转的节点设为start，开始一个一个将start后面的node接到pre的后面。

代码：
recursive：

public ListNode reverseList(ListNode head) {
    if(head == null || head.next == null){
        return head;
    }
    ListNode next = head.next;
    ListNode newHead = reverseList(next);
    next.next = head;
    head.next = null;
    return newHead;
}

iterative:

public ListNode reverseList(ListNode head) {
    if(head == null || head.next == null){
        return head;
    }
    ListNode curr = head;
    ListNode next = curr.next;
    curr.next = null;
    while(next != null){
        ListNode nextnext = next.next;
        next.next = curr;
        curr = next;
        next = nextnext;
    }
    return curr;
}

follow up：

public ListNode reverseBetween(ListNode head, int m, int n) {
    if(head == null) return null;
    ListNode dummy = new ListNode(0); // create a dummy node to mark the head of this list    dummy.next = head;
    ListNode pre = dummy; // make a pointer pre as a marker for the node before reversing    for(int i = 0; i<m-1; i++) pre = pre.next;

    ListNode start = pre.next; // a pointer to the beginning of a sub-list that will be reversed    ListNode then = start.next; // a pointer to a node that will be reversed
    // 1 - 2 -3 - 4 - 5 ; m=2; n =4 ---> pre = 1, start = 2, then = 3    // dummy-> 1 -> 2 -> 3 -> 4 -> 5
    for(int i=0; i<n-m; i++)
    {
        start.next = then.next;
        then.next = pre.next;
        pre.next = then;
        then = start.next;
    }

    // first reversing : dummy->1 - 3 - 2 - 4 - 5; pre = 1, start = 2, then = 4    // second reversing: dummy->1 - 4 - 3 - 2 - 5; pre = 1, start = 2, then = 5 (finish)
    return dummy.next;

}