Question

我有一种方法可以交换循环链表的两个节点

它使用2个对象作为参数。该代码适用于以下情况这两个对象都是头部的元素

public void exchange(E a, E b) {
 Node<E> head = tail.getNext();
 if(a == b)
     return;

 if(isEmpty() || size == 1) //no exchange if size is 1
     return;

 if(size == 2){   //if size is two, head will be tail and tail will be head
     rotate();
 }

 else{

    Node<E> currA = head, prevA = null;    //search for first node
         if(head.getElement() == a){
         Node<E> temp = head;
         while(!temp.getNext().equals(head)){
             temp = temp.getNext();
         }
         prevA = temp;

     }else if(tail.getElement() == a){
        currA = tail;
        while(!currA.getNext().equals(tail)){
            prevA = currA.getNext();
        }
    }else{
        while(!currA.getElement().equals(a)){
            prevA = currA;
            currA = currA.getNext();
            if(currA.getNext() == head)
                break;
        }
    }

    Node<E> currB = head, prevB = null; //Search for second node
              if(currB.getElement() == b){
         Node<E> temp = head;
     while(!temp.getNext().equals(head)){
         temp = temp.getNext();
     }
     prevB = temp;

     }else if(tail.getElement() == b){
         currB = tail;
        while(!currB.getNext().equals(tail)){
            prevB = currB.getNext();
        }
     }else{
         while(!currB.getElement().equals(b)){
             prevB = currB;
             currB = currB.getNext();
             if(currB.getNext() == head)
                 break;
     }
 }
 System.out.println(currA.getElement() +" currentA"); //I get output for this
 System.out.println(currB.getElement() +" currentB");
 System.out.println(prevA.getElement() +" previousA");// Gives null pointer exception when object a is head 
 System.out.println(prevB.getElement() +" previousB");// Gives null pointer exception when object b is head
 prevA.setNext(currB);
 prevB.setNext(currA);
 Node<E> temp = currA.getNext();
 currA.setNext(currB.getNext());
 currB.setNext(temp);
 }

}

我的代码能够提供和使用currA（和currB）
但是
如果a与head相同，则它不是空指针，而不是在if语句中分配prevA。如果b是head的元素，则发生相同的情况，则prevB为null。（代码在此处停止）

Java循环链表交换不适用于头部

0 个答案: