题目
输入一个链表的头结点,从尾到头反过来打印出每个结点的值。结点定义如下:
public class ListNode {
int val;
ListNode next = null;
ListNode(int val) {
this.val = val;
}
}
思路
结点遍历顺序只能从头到尾,但是输出的顺序却为从尾到头,是典型的“后进先出”问题,这就要联想到使用栈,从而也可以联想到使用递归。
测试用例
1.功能测试(单个结点链表,多个结点链表)
2.特殊输入测试(链表为空)
完整Java代码
public class PrintListInReversedOrder {
class ListNode{
int key;
ListNode next;
public ListNode(int key) {
this.key=key;
this.next=null;
}
}
// 采用栈
public void printListReversingly_Iteratively(ListNode node) {
Stack<ListNode> stack = new Stack<ListNode>();
while (node!= null) {
stack.push(node);
node=node.next;
}
while(!stack.empty()) {
System.out.println(stack.pop().key);
}
}
//采用递归
public void printListReversingly_Recursively(ListNode node) {
if(node!=null) {
printListReversingly_Recursively(node.next);
System.out.println(node.key);
}else
return;
}
// ==================================测试代码==================================
/**
* 链表为空
*/
public void test1() {
ListNode aListNode = null;
System.out.println("采用栈:");
printListReversingly_Iteratively(aListNode);
System.out.println("采用递归:");
printListReversingly_Recursively(aListNode);
}
/**
* 多个结点链表
*/
public void test2() {
ListNode ListNode1 = new ListNode(1);
ListNode ListNode2 = new ListNode(2);
ListNode ListNode3 = new ListNode(3);
ListNode ListNode4 = new ListNode(4);
ListNode ListNode5 = new ListNode(5);
ListNode1.next=ListNode2;
ListNode2.next=ListNode3;
ListNode3.next=ListNode4;
ListNode4.next=ListNode5;
System.out.println("采用栈:");
printListReversingly_Iteratively(ListNode1);
System.out.println("采用递归:");
printListReversingly_Recursively(ListNode1);
}
/**
* 单个结点链表
*/
public void test3() {
ListNode ListNode1 = new ListNode(1);
System.out.println("采用栈:");
printListReversingly_Iteratively(ListNode1);
System.out.println("采用递归:");
printListReversingly_Recursively(ListNode1);
}
public static void main(String[] args) {
PrintListInReversedOrder demo = new PrintListInReversedOrder();
System.out.println("test1:");
demo.test1();
System.out.println("test2:");
demo.test2();
System.out.println("test3:");
demo.test3();
}
}