我想在java中反转单链表中的偶数,但要获得正确的输出却面临一些困难。例如,input:2、18、24、3、5、7、9、6、12方法应该只反转偶数,即{2,18,24}和{6,12}
正确的输出:24、18、2、3、5、7、9、12、6
但是,我的输出:24 18 3 5 7 9 12 6哪个是错误的
主要方法
public static void main(String[] args) throws Exception {
SLL<Integer> p = new SLL<Integer>();
int[] e = { 2, 18, 24, 3, 5, 7, 9, 6, 12,5 ,4 ,3 ,2,6,8};
for (int i = 0; i < e.length; i++) {
p.addToHead(e[i]);
}
p = reverse(p);
p.printAll();
}
这就是方法(不能正常工作)
public static SLL<Integer> reverse(SLL<Integer> p) {
SLL<Integer> returnList = new SLL<Integer>();
Stack<Integer> stk = new Stack<Integer>();
for (SLLNode tmp = p.getHead(); tmp != null; tmp = tmp.next) {
if ((((Integer) tmp.info) % 2) != 0) {
returnList.addToHead((Integer) tmp.info);
p.deleteFromHead();
} else if ((((Integer) tmp.info) % 2) == 0) {
stk.push((Integer) tmp.info);
p.deleteFromHead();
}
if (stk.getLSize() >= 2) {
while (!(stk.isEmpty())) {
returnList.addToHead((Integer) stk.pop());
}
}
}
return returnList;
}
这是sllnode类
public class SLLNode<T> {
public T info;
public SLLNode<T> next;
public SLLNode() {
this(null,null);
}
public SLLNode(T el) {
this(el,null);
}
public SLLNode(T el, SLLNode<T> ptr) {
info = el;
next = ptr;
}
}
这是sll类
public class SLL<T> {
protected SLLNode<T> head, tail;
public SLL() {
head = tail = null;
}
public boolean isEmpty() {
return head == null;
}
public void addToHead(T el) {
head = new SLLNode<T>(el, head);
if (tail == null)
tail = head;
}
public SLLNode getHead(){
return head;
}
public void addToTail(T el) {
if (!isEmpty()) {
tail.next = new SLLNode<T>(el);
tail = tail.next;
} else
head = tail = new SLLNode<T>(el);
}
public T deleteFromHead() { // delete the head and return its info;
if (isEmpty())
return null;
T el = head.info;
if (head == tail) // if only one node on the list;
head = tail = null;
else
head = head.next;
return el;
}
public T deleteFromTail() { // delete the tail and return its info;
if (isEmpty())
return null;
T el = tail.info;
if (head == tail) // if only one node in the list;
head = tail = null;
else { // if more than one node in the list,
SLLNode<T> tmp; // find the predecessor of tail;
for (tmp = head; tmp.next != tail; tmp = tmp.next)
;
tail = tmp; // the predecessor of tail becomes tail;
tail.next = null;
}
return el;
}
public void delete(T el) { // delete the node with an element el;
if (!isEmpty())
if (head == tail && el.equals(head.info)) // if only one
head = tail = null; // node on the list;
else if (el.equals(head.info)) // if more than one node on the list;
head = head.next; // and el is in the head node;
else { // if more than one node in the list
SLLNode<T> pred, tmp;// and el is in a nonhead node;
for (pred = head, tmp = head.next; tmp != null
&& !tmp.info.equals(el); pred = pred.next, tmp = tmp.next)
;
if (tmp != null) { // if el was found;
pred.next = tmp.next;
if (tmp == tail) // if el is in the last node;
tail = pred;
}
}
}
public void printAll() {
for (SLLNode<T> tmp = head; tmp != null; tmp = tmp.next)
System.out.print(tmp.info + " ");
}
public boolean isInList(T el) {
SLLNode<T> tmp;
for (tmp = head; tmp != null && !tmp.info.equals(el); tmp = tmp.next)
;
return tmp != null;
}
public int length() {
int length = 0;
for (SLLNode tmp = head; tmp != null; tmp = tmp.next) {
length += 1;
}
return length;
}
1条答案
按热度按时间qf9go6mv1#
运行你的反向方法给我一个与你看到的不同的输出。所以,我怀疑你的输出来自稍微不同的代码。
我得到:24,6,9,7,5,3,2,18
在反向方法中,当堆栈上有2个时,就开始向returnlist添加偶数。如果要反转所有偶数,则需要等待堆栈上的所有连续偶数。或者换句话说,当你得到一个奇数,或者没有剩下的数,你可以把所有的偶数从堆栈中弹出。。
我认为你也应该用addtail而不是addhead。
比如说