跬步

比较常见的一道算法题，使用额外空间使此问题变得十分简单，本文记录一下不使用额外空间的两种方法。

链表类

public class ListNode {
    Integer value;
    ListNode next;
    
    public ListNode(Integer value){
        this.value = value;
    }
    
    @Override
    public String toString(){
        StringBuilder str = new StringBuilder(this.value == null?"null":this.value.toString());
        ListNode cur = this;
        while (cur.next != null){
            str = str.append("->").append(cur.next.value);
            cur = cur.next;
        }
        return str.toString();
    }
}

我重写了一下toString方法，比较方便看结构。

循环方法

    public static ListNode reverseLoop(ListNode node){
        ListNode prev = null;
        ListNode head = node;
        ListNode next;
        while (head != null) {
            next = head.next;
            head.next = prev;
            prev = head;
            head = next;
        }
        return prev;
    }

递归方法

    public static ListNode reverseIter(ListNode head){
        if(head == null || head.next == null) {
            return head;
        }
        ListNode result = reverseIter(head.next);
        head.next.next = head;
        head.next = null;
        return result;
    }

测试

    public static void main(String[] args){
        ListNode node1 = new ListNode(1);
        ListNode node2 = new ListNode(2);
        ListNode node3 = new ListNode(3);
        ListNode node4 = new ListNode(4);
        ListNode node5 = new ListNode(5);
        ListNode node6 = new ListNode(6);
        ListNode node7 = new ListNode(7);
        ListNode node8 = new ListNode(8);
        ListNode node9 = new ListNode(null);
        node1.next = node2;
        node2.next = node3;
        node3.next = node4;
        node4.next = node5;
        node5.next = node6;
        node6.next = node7;
        node7.next = node8;
        node8.next = node9;
        System.out.println(reverseLoop(node1));
        System.out.println(reverseIter(node9));
    }

输出结果

null->8->7->6->5->4->3->2->1
1->2->3->4->5->6->7->8->null