Reverse a singly linked list.

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Hint:
A linked list can be reversed either iteratively or recursively. Could you implement both?

iterative 解法：

总结就是得到下一个节点，更改当前节点指向，将指针往下移动，直到过完整个linkedlist.

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode reverseList(ListNode head) {
        ListNode prev = null;
        ListNode cur = head;
        while(cur!=null){
            ListNode next = cur.next; 
            cur.next = prev;
            prev = cur;
            cur = next;
        }
        return prev;
    }
}
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode reverseList(ListNode head) {
        if(head == null || head.next==null)
            return head;
        // 1->2->3->4->null   
        ListNode nextNode = head.next;          // 2->3->4->null 
        ListNode reversed = reverseList(nextNode);  //4->3->2->null 
        nextNode.next = head;      //4->3->2->1 
        head.next = null;               //4->3->2->1->null
        return reversed;      
    }
}
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode reverseList(ListNode head) {
        ListNode rev = null;
        ListNode temp1 = head;
        while(temp1 != null) {
            if (rev == null) {
                rev = new ListNode(temp1.val);
            } else {
                ListNode tnew = new ListNode(temp1.val);
                tnew.next = rev;
                rev = tnew;
            }
            temp1 = temp1.next;
        }
        return rev;
    }
}