Question

这对于学校作业可能有人指出我正确的方向。我试图使用递归在odds和evens之间拆分链接列表，但是.add（IntNode）似乎有些错误，它在前2个节点之间保持交替，就像它覆盖它一样。我已经解决了我能想到的一切，似乎我使用添加不正确。我必须编辑赔率和均衡方法，并将原始链表保留为空。

这是我的代码，任何帮助都将不胜感激。

测试代码：

public class Test1 {

    public static void main(String[] args) { 
        IntLinkedList test = new IntLinkedList(new int[]{3,5,4,6,8,2,1,4,3,2,5,3,6,3});
        System.out.println("Test list: "+test);

        IntLinkedList oddOnes = new IntLinkedList(), evenOnes = new IntLinkedList();
        test.split(oddOnes,evenOnes);
        System.out.println("Test list should be empty: "+test);
        System.out.println("Odds: "+oddOnes);
        System.out.println("Evens: "+evenOnes);
    }

}

IntLinkedList相关方法：

public void add(IntNode node) { 
    if (first != null) {
        node.setNext(first);
        first = node;
    } else {
        first = node;
    }
}

public void split(IntLinkedList odds, IntLinkedList evens) {
    IntNode curr = first;

     if (curr != null) {
        if (curr.getData() % 2 == 0) {
            evens.add(curr);
        } else {
            odds.add(curr);
        }
        System.out.println(curr.getData());
        remove(0);
        split(odds, evens);
    }
}

public void remove(int index) {
    IntNode curr = first;
    IntNode past = null;
    int currIndex = 0;

    while (curr != null && currIndex < index) {
        past = curr;
        curr = curr.getNext();
        currIndex++;
    }

    if (past == null && curr != null && index > -1) {
        first = first.getNext();
        curr.setNext(null);
    } else if (curr != null && index > -1) {
        past.setNext(curr.getNext());
    }
}

IntNode：

public class IntNode {

    private int number;
    private IntNode nextInt;

    public IntNode(int number) {
        this.number = number;
    }

    public IntNode(int number, IntNode next) {
        this.number = number;
        this.nextInt = next;
    }

    public IntNode getNext() {
        return nextInt;
    }

    public void setNext(IntNode nextInt) {
        this.nextInt = nextInt;
    }

    public int getData() {
        return number;
    }

}

Answer 1

在将当前节点添加到奇数或偶数列表之前，必须先将其删除。否则，您将在“添加”方法调用中修改源列表。

System.out.println(curr.getData());
remove(0);
if (curr.getData() % 2 == 0) {
    evens.add(curr);
} else {
    odds.add(curr);
}
split(odds, evens);

如何将现有节点添加到链接列表的开头？

1 个答案: