算法练习(36): 链表的增删查改2(1.3.21-1.3.23

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算法(第4版)

知识点

• 链表节点增删查改

题目

1.3.21 编写一个方法find()，接受一条链表和一个字符串key作为参数。如果链表中的某个结点的item域的值为key，则方法返回true，否则返回false。

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1.3.21 Write a method find() that takes a linked list and a string key as arguments and returns true if some node in the list has key as its item field, false otherwise.

答案

public class LinkedListExecise3 {

    private static class Node {
        Node next;
        String item;
    }

    public boolean find(Node first, String key) {
        if (first == null)
            return false;
        Node current = first;
        while (current != null) {
            if (current.item.equals(key)) {
                return true;
            }
            current = current.next;
        }

        return false;
    }

    /**
     * @param args
     */
    public static void main(String[] args) {
        // TODO Auto-generated method stub

        /**
         * 创建链表
         * */
        Node first = new Node();
        Node second = new Node();
        Node third = new Node();
        Node forth = new Node();
        Node fifth = new Node();
        first.item = "我的";
        first.next = second;
        second.item = "名字";
        second.next = third;
        third.item = "叫";
        third.next = forth;
        forth.item = "顶级程序员不穿女装";
        forth.next = fifth;
        fifth.item = "微博:https://m.weibo.cn/p/1005056186766482";
        fifth.next = null;

        String targetString = "微博:https://m.weibo.cn/p/1005056186766482";
        LinkedListExecise3 linkedListExecise3 = new LinkedListExecise3();
        boolean find = linkedListExecise3.find(first, targetString);
        System.out.println("查找字符串" + targetString + ":" + find);

    }

代码索引

LinkedListExecise3.java

题目

1.3.22 假设x是一条链表中的某个结点，下面这段代码做了什么？

t.next = x.next;
x.next = t;

答案：将节点t插入到结点x后面。

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**1.3.22 Suppose that x is a linkedlist Node.What does the following code fragment do? **

t.next = x.next;
x.next = t;

Answer : Inserts node t immediately after node x.

题目

1.3.23 为什么下面这段代码和上一道题中的代码效果不同？

x.next = t;
t.next = x.next;

在更新t.next时，x.next已经不再指向x的原来后续结点，而是指向t本身。

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1.3.23 Why does the following code fragment not do the same thing as in the previous question?

x.next = t;
t.next = x.next;

Answer : When it comes time to update t.next, x.next is no longer the original node following x, but is instead t itself!

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