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java左右旋转列表

我正在考虑从列表中获取下一个和上一个元素,但我无法正确处理此问题-问题发生在列表的限制上。也许有人有更好的概念来解决这个问题

public class Test {

    private char heading = 'N';
    List<Character> cardinal = Arrays.asList('N', 'E', 'S', 'W' );
    ListIterator<Character> iterator = cardinal.listIterator();

    public static void main(String[] args) {

        Test test = new Test();
        System.out.println("heading = " + test.heading);

        test.rotateLeft(); System.out.println("Rotating left...");
        System.out.println("heading = " + test.heading);
        test.rotateLeft(); System.out.println("Rotating left...");
        System.out.println("heading = " + test.heading); 
   }

    public void rotateRight() {
        this.heading =
            (iterator.hasNext() ? iterator.next() : cardinal.get(0));
    }  

    public void rotateLeft() {
        this.heading = 
                (iterator.hasPrevious() ? iterator.previous() : cardinal.get(3));
    }

}

共 (1) 个答案

  1. # 1 楼答案

    问题是,当到达列表末尾时,没有重置迭代器。从那时起,iterator.hasNext()将始终返回false。一个简单的解决方法:

    public void rotateRight() {
        if (!iterator.hasNext())
            iterator = cardinal.listIterator();
        this.heading = iterator.next();
    }
    
    public void rotateLeft() {
        if (!iterator.hasPrevious())
            iterator = cardinal.listIterator(cardinal.size());
        this.heading = iterator.previous();
    }
    

    如果您的列表为final,则更容易:

    int index = 0;
    
    public void rotateRight() {
        index = (index + 1) % cardinal.size();
        this.heading = cardinal.get(index);
    }
    
    public void rotateLeft() {
        index = (index - 1 + cardinal.size()) % cardinal.size();
        this.heading = cardinal.get(index);
    }