单链表反转

输入一个链表，反转链表后，输出新链表的表头。

 

第一种用了四个指针，没有下一个简单

public class Solution {
    public ListNode ReverseList(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }
        ListNode cur = head;
        ListNode p = head.next;


        while (p != null) {
            ListNode q = p.next;
            p.next = cur;
            cur = p;
            p = q;
        }
        head.next = null;
        return cur;
    }
}

class ListNode {
    int val;
    ListNode next = null;

    ListNode(int val) {
        this.val = val;
    }
}

 

这个解发用了三个，比较好

public class Solution {
    public ListNode ReverseList(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }

        ListNode cur = head.next;
        head.next = null;//第一个节点变成尾节点，指向null

//        下面是大于1个节点的情况
        while (cur != null) {

            ListNode temp = cur.next;
            cur.next = head;
            head = cur;
            cur = temp;
        }
        return head;
    }
}

class ListNode {
    int val;
    ListNode next = null;

    ListNode(int val) {
        this.val = val;
    }
}