使用linkedList的mergesort算法无法在Java中正确实现
我已经为一个自定义的LinkedIntList
实现了mergesort方法,它只不过是一个LinkedList
类。我很难理解它为什么失败。也就是说,它不会产生任何结果。我把代码检查了好几次,都弄不明白
代码如下:
class ListNode {
public int data; // data stored in this node
public ListNode next; // link to next node in the list
// post: constructs a node with data 0 and null link
public ListNode() {
this(0, null);
}
// post: constructs a node with given data and null link
public ListNode(int data) {
this(data, null);
}
// post: constructs a node with given data and given link
public ListNode(int data, ListNode next) {
this.data = data;
this.next = next;
}
}
LinkedIn列表类:
public class LinkedIntList {
public ListNode front;
// Constructs an empty list.
public LinkedIntList() {
this(null);
}
public LinkedIntList(ListNode node){
front=node;
}
public String toString() {
if (front == null) {
return "[]";
} else {
String result = "[" + front.data;
ListNode current = front.next;
while (current != null) {
result += ", " + current.data;
current = current.next;
}
result += "]";
return result;
}
public void sort(){
front=mergeSort(front);
}
public ListNode mergeSort(ListNode node) {
//ystem.out.println("merge sort is called");
//first get middle of linkedlist, using two pointers
//you can also get this by size/2
//step 1: get middle pointers
if (node==null || node.next==null)
return null;
ListNode runner=node.next.next;
ListNode walker=node;
while(runner!=null && runner.next!=null){
runner=runner.next.next;
walker=walker.next;
}
//At this point walker is in center
ListNode right=walker.next;
walker.next=null;
ListNode left=node;
left=mergeSort(left);
right=mergeSort(right);
node=merge(left,right);
return node;
}
//merge of two linkedlist happens here
private ListNode merge(ListNode l1, ListNode l2) {
if (l1==null){
return l2;
}
if (l2==null){
return l1;
}
ListNode head=null;
ListNode curr=null;
ListNode temp=null;
while(l1!=null && l2!=null){
temp=(l1.data < l2.data ? l1 : l2);
if (head==null){
head=temp;
curr=head;
}
else {
curr.next=temp;
curr=curr.next;
}
if (temp==l1){
l1=l1.next;
}
else l2=l2.next;
}
if (l1!=null){
curr.next=l1;
}
if (l2!=null){
curr.next=l2;
}
return head;
}
}
测试此功能的实用程序类:
public class UtilityMain {
public static void main(String[] args){
ListNode node5 = new ListNode(1,null);
ListNode node4=new ListNode(2,node5);
ListNode node3=new ListNode(3,node4);
ListNode node2=new ListNode(4,node3);
ListNode node1 = new ListNode(5,node2);
LinkedIntList l = new LinkedIntList(node1);
System.out.println("before sorting " + l);
l.sort();
System.out.println("Afer sorting " + l);
}
}
输出:--->
before sorting [5, 4, 3, 2, 1]
After sorting []
# 1 楼答案
至少部分问题在这里:
如果当前节点为null,则应返回null, 但是,如果节点。如果next为null,则应返回node
考虑列表(3, 2)
的情况列表节点为(节点[3],下一个->;节点[2],下一个->;[NULL])
您可以使用Walker=(节点[3],Next->;[NULL])和Runner=(节点[2],Next->;[NULL])来打破列表
然后在Walker(重命名为左侧)和Runner(重命名为右侧)上调用合并排序
它位于这些调用的每个节点上。下一个测试为null,因此它返回null,而不是您想要的列表,它应该是传入的列表
存在一个错误,该错误将在合并方法中生成空指针异常。。你能找到它吗