java - java获取链表中的最小元素

Question

我正在尝试从链接列表中获取最小元素。然而我的当前代码并没有遍历所有元素，它只检查第一对元素。我知道我的错误在哪里，只是不知道为什么它没有遍历，即使我使用了 next。

    public class ListOfNVersion03PartB
{   
    private int thisNumber;              // the number stored in this node
    private ListOfNVersion03PartB next;  // forms a linked list of objects

    private final int nodeID;            // a unique ID for each object in the list

    private static int nodeCount = 0;    // the number of list objects that have been created


    public ListOfNVersion03PartB(int num)
    {
        thisNumber = num;
        next = null;

        ++nodeCount;
        nodeID = nodeCount;

    } 

public ListOfNVersion03PartB(int [] num)
{
    this(num[0]);  // in this context, "this" invokes the other constructor

    for (int i=1 ; i<num.length ; ++i)
        insertLast(num[i]);

} 

        public int minVal()
    {

        if(next.thisNumber> thisNumber)
        return thisNumber;

        else
         return next.minVal();   

    }

Answer 1

因为你没有将next分配给另一个节点，并且它的值为null，所以当你调用minVal()时会抛出NullPointerException

您可以检查 next 是否为 null，如下所示

        public int minVal() {
            ListOfNVersion03PartB current = this;
            int min = Integer.MAX_VALUE;
            while (current != null) {
                if (current.thisNumber < min)
                    min = current.thisNumber;
                current = current.next;
            }
            return min;
        }