[英]mergesort using linkedList not working correctly - implemented in Java
我已经为自定义LinkedIntList
实现了mergesort方法,它只是一个LinkedList
类。 我很难理解它失败的原因。 那就是它不会产生任何结果。 我多次通过代码,我无法弄清楚。
这是代码:
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;
}
}
LinkedIntList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;
}
}
测试这个的实用程序类:
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 []
至少部分问题在于:
if (node==null || node.next==null)
return null;
如果当前节点为null,则应返回null,但是,如果node.next为null,则应返回node。
考虑列表的情况[3,2]
列表节点是(NODE [3],Next - > NODE [2],Next - > [NULL])
你用Walker =(NODE [3],Next - > [NULL])和Runner =(NODE [2],Next - > [NULL])打破列表
然后在Walker(重命名为Left)和Runner(重命名为Right)上调用合并排序
在每个调用中,Node.next测试为null,因此它返回null而不是您想要的列表,这应该是传入的列表。
有一个错误会在合并方法中生成一个NULL指针异常。你能找到吗?
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