双重链表的递归方法

Question

我需要开发一个递归方法(不能使用while, do…如果i>>= 0且i小于链表的值，则返回链表的第i个元素。否则返回null。任何帮助都将非常感激。这是我的迭代方法:

public String get(int i) {
    if(i<0 || i>=lenght) {
        return null;
    }
    Node t = head;
    for(int c = 0; c != i; c++) {
       t = t.next;
    }
    return t.element;
}

Answer 1

public String get(Node current, int currentIndex, int targetIndex) {
    // first check the exit condition of the method
    if(currentIndex <0 || currentIndex >= length || current == null) {
            return null;
    }
    // check the second exit contidion
    if (currentIndex == targetIndex)
    {
        return current.element;
    }
    // go forward by increasing the index
    return get(current.next, currentIndex + 1, targetIndex);
} 

Answer 2

试试这样:

public String get(int i) {
    if(i<0 || i>=lenght) return null;
    return iter(head, i)
}

public String iter(Node t,int i){
    if(t == null) return null;
    if(i == 0) return t.elemnt;
    return iter(t.next, i - 1)
}