这里,我想以递归方式反转链表的每个k元素。对于链表
3 → 4 → 5 → 2 → 6 → 1 → 9
的{{1}}变为kReverse(3)
我收到5 → 4→ 3→ 1→ 6→ 2→ 9→ 1
。
NullPointerException
答案 0 :(得分:2)
粗略地看一下,您不会更新递归以使用newHead。
另外,你需要注意打破和制作指针。当你保留前3个元素时,他们需要指向前一个元素。
4.next = 3和5.next = 4(你需要在代码中管理它)
一个提示:
使用大小为k的堆栈可以更容易地反转所需的元素并在其完全递归时弹出。
答案 1 :(得分:0)
我不确定这是否是你想要的? 不需要大小,
public class Node<T> {
private T item;
private Node<T> next;
public Node(T item, Node<T> next) {
this.item = item;
this.next = next;
}
public Node(T item) {
this.item = item;
}
public Node<T> getNext() {
return next;
}
public void setNext(Node<T> next) {
this.next = next;
}
public static Node<Integer> reverseLinkedList(Node<Integer> head) {
if (head.next != null){
Node<Integer> prev = head.next;
Node<Integer> result = reverseLinkedList(prev);
prev.next = head;
return result;
} else {
return head;
}
}
private static Node<Integer> kReverseLinkedList(Node<Integer> head, Node<Integer> cursor, int counter, int k) {
Node<Integer> result;
if (cursor.next == null){
result = reverseLinkedList(head);
head.next = null;
} else if (counter > 0){
result = kReverseLinkedList(head, cursor.next, counter-1, k);
} else {
Node<Integer> next = cursor.next;
cursor.next = null;
result = reverseLinkedList(head);
head.next = kReverseLinkedList(next, next, k, k);
}
return result;
}
public static Node<Integer> kReverseLinkedList(Node<Integer> head, int k) {
Node<Integer> result = null;
if (head != null){
result = head;
if (k > 1) {
result = kReverseLinkedList(head, head, k-1, k-1);
}
}
return result;
}
public static void print(Node<Integer> n) {
if (n != null){
System.out.print(n.item+" ");
print(n.next);
} else {
System.out.println();
}
}
public static void main(String[] args) {
int[] data = {3,4,5,2,6,1,9};
Node<Integer> head = new Node<>(data[0]);
Node<Integer> tail = head;
for (int i = 1; i < data.length; i++) {
Node<Integer> n = new Node<>(data[i]);
tail.setNext(n);
tail = n;
}
print(head);
head = kReverseLinkedList(head, 3);
print(head);
}
}
输出:
3 4 5 2 6 1 9
5 4 3 1 6 2 9
答案 2 :(得分:0)
我认为你不需要在递归版本中使用循环。
要使其通用,请添加偏移。
不需要尺寸。
这是一个带偏移(未经测试)的真实递归版本:
private static Node<Integer> tail;
public static Node<Integer> kReverseLinkedList(Node<Integer> head, int k, int offset, Node<Integer> current) {
if (offset > 1) {
kReverseLinkedList(head.next, k, offset - 1, head.next);
return head;
} else if (offset == 1) {
head.next = kReverseLinkedList(head.next, k, 0, head.next);
return head;
} else if (k == 1) {
tail = current.next;
return current;
} else {
Node<Integer> n = kReverseLinkedList(head, k - 1, 0, current.next);
current.next.next = current;
if (current == head)
head.next = tail;
return n;
}
}
// usage: kReverseLinkedList(myListHead, 8, 2, myListHead); (k >= 2, offset >= 0)