[剑指offer-6]从尾到头打印链表

最新推荐文章于 2023-06-22 17:10:40 发布

原创最新推荐文章于 2023-06-22 17:10:40 发布 · 90 阅读

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CC 4.0 BY-SA版权

本文介绍了一种在Java和Python中实现链表逆序打印的方法，包括使用栈和递归两种方式。通过具体实例展示了如何创建链表并逆序打印其结点值。

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题目描述
输入个链表的头结点，从尾到头反过来打印出每个结点的值
java版

 public static class ListNode {
        int val; //节点的值
        ListNode next; //下一个节点的值

        public ListNode() {
        }

        public ListNode(int val) {
            this.val = val;
        }
    }
    //使用栈
    public static void printListInverselyUsingStack(ListNode head) {
        Stack<ListNode> stack = new Stack<>();
        //入栈
        while (head != null) {
            stack.push(head);
            head = head.next;
        }
        //出栈
        ListNode temp;
        while (!stack.empty()) {
            temp = stack.pop();
            System.out.println(temp.val+" ");
        }
    }
    //使用递归
    public static void printListInverselyUsingIteration(ListNode head){
        if (head != null){
            printListInverselyUsingStack(head.next);
            System.out.println(111);
            System.out.println(head.val);
        }

    }
    public static void main(String[] args) {
        ListNode root = new ListNode();
        root.val=1;
        root.next = new ListNode(2);
        root.next.next = new ListNode(3);
        root.next.next.next = new ListNode(4);
        root.next.next.next.next = new ListNode(5);

//        printListInverselyUsingStack(root);
//        printListInverselyUsingStack(null);
        printListInverselyUsingIteration(root);
    }

python版

class ListNode:
    def __init__(self,x=None):
        self.val = x;
        self.next = None
        
class solution:
    def printListInverse(self,listNode):
        if listNode == None:
            return 
        temp = []
        head = listNode
        while head:
            temp.insert(0, head.val)
            head = head.next
            
        return temp
    
node1=ListNode(10)
node2=ListNode(20)
node3=ListNode(30)

SimpleNode = ListNode(2)
test = ListNode()

node1.next = node2
node2.next = node3

s = solution()
print(s.printListInverse(node1))
print(s.printListInverse(test))
print(s.printListInverse(SimpleNode))