鏈表實現排序方法

Question

我不是從Java集合中創建鏈接列表，而是將其放入許多方法中..我的鏈接列表的工作方式與原始列表相同，但是我的類具有打印和排序 方法，但有一些細微差別， 我的輸入是：[5,18,3 ,, 10,2] ...或它可以是字符串輸入 ..我想使用sort方法對鏈表進行排序，因此輸出應為：[2,3,5,10,18]或有序字符串是排序方法代碼：

        public void sort(){
        Node<E> current = head ;
        Node<E> current2 = current.next;
        E min = head.element;
        E temp;
        int pos = 0;
        for (int i = 0; i < size-1; i++) { 
            for (int j = 0; j < size; j++) {
                if(current2 != null){
                    if(min.compareTo(current2.element) > 0){
                      pos = j ;
                      min = current2.element;

                    }
                    current2= current2.next;
                }
            }

            temp = current.element;
            current.element = min;
            current = current.next;
            min = current.element;
            current2 = head;
            for (int j = 0; j <= pos; j++) {
                if(current2 !=null){
                if(j==pos){current2.element = temp;}
                current2= current2.next;
                } 
            }
            current2 = current.next;   
    }
}

完整的代碼

Answer 1

嘗試使用這種排序方法。

public void sort()
  {
    for (int i = size - 1; i >= 1; i--)
    {
      Node<E> finalNode = head;
      Node<E> tempNode = head;

      for (int j = 0; j < i; j++)
      {
        E val1 = head.element;
        Node<E> nextnode = head.next;
        E val2 = nextnode.element;
        if (val1.compareTo(val2))
        {
          if (head.next.next != null)
          {
            Node<E> CurrentNext = head.next.next;
            nextnode.next = head;
            nextnode.next.next = CurrentNext;
            if (j == 0)
            {
              finalNode = nextnode;
            }
            else
              head = nextnode;

            for (int l = 1; l < j; l++)
            {
              tempNode = tempNode.next;
            }

            if (j != 0)
            {
              tempNode.next = nextnode;

              head = tempNode;
            }
          }
          else if (head.next.next == null)
          {
            nextnode.next = head;
            nextnode.next.next = null;
            for (int l = 1; l < j; l++)
            {
              tempNode = tempNode.next;
            }
            tempNode.next = nextnode;
            nextnode = tempNode;
            head = tempNode;
          }
        }
        else
          head = tempNode;
        head = finalNode;
        tempNode = head;
        for (int k = 0; k <= j && j < i - 1; k++)
        {
          head = head.next;
        }

      }
    }
  }