我们在课堂上已经看到选择排序算法如何处理数组数据结构。在本实验中，我们将练习如何对链表ADT执行选择排序。 1.转换以下选择排序伪代码，以升序执行排序。 （selectionSort_asc函数） 一种。在长度为n的链表中找到最小值的节点 b。将min节点追加到新的结果链接列表中 C。从原始链接列表中删除分钟 d。重复步骤a-c，直到原始链表为空 e。返回结果链接列表 2.转换以下选择排序伪代码，以降序执行排序。 （selectionSort_desc函数） 一种。在长度为n的链表中找到最大值的节点 b。将最大节点追加到新的结果链接列表中 C。从原始链表中删除最大值 d。重复步骤a-c，直到原始链表为空 e。返回结果链接列表
我在下面尝试了此代码，但没有给我正确的输出。
public class Sort {
public static void main(String[] args){
Node head = initializeList("ilovedata");
System.out.println("\n List Before selectionSort_asc");
printList(head);
head = selectionSort_asc(head);
System.out.println("\n List After selectionSort_asc");
printList(head);
// Expected answer: -> a -> a -> d -> e -> i -> l -> o -> t -> v
head = initializeList("ilovedata");
System.out.println("\n List Before selectionSort_desc");
printList(head);
head = selectionSort_desc(head);
System.out.println("\n List After selectionSort_desc");
printList(head);
// Expected answer: -> v -> t -> o -> l -> i -> e -> d -> a -> a
}
public static Node selectionSort_asc(Node head){
Node result = null;
Node curr, prev, min;
while(head!=null) {
curr = head;
prev = null;
min = head;
while(curr.next!=null) {
curr = curr.next;
if(curr.item<min.item) {
prev = min;
min = curr;
}
}
//append the min at the end of result list
Node add_min = new Node(min.item);
if(result==null)
result = add_min;
else {
Node last = result;
while(last.next!=null) {
last = last.next;
}
last.next = add_min;
}
//delete the min node form original list
if(min==head) {
head = head.next;
}
else if(min.next==null){
prev.next = null;
}else {
prev.next = prev.next.next;
min.next = null;
}
}
return result;
}
public static Node selectionSort_desc(Node head){
Node result = null;
Node curr, prev, max;
while(head!=null) {
curr = head;
prev = null;
max = head;
//find out the max node
while(curr.next!=null) {
curr = curr.next;
if(curr.item>max.item) {
prev = max;
max = curr;
}
}
//add max to the end of result list
Node add_max = new Node(max.item);
if(result==null) {
result = add_max;
}
else {
Node last = result;
while(last.next!=null) {
last = last.next;
}
last.next = add_max;
}
//delete min from original list
if(max == head) {
head = head.next;
}
else if(max.next==null){
prev.next = null;
}
else {
prev.next = max.next;
max.next = null;
}
}
return result;
}
// Method that takes a string and insert its characters into a linked list
public static Node initializeList(String str){
Node head= new Node(str.charAt(0)),cur;
int i;
for(cur= head,i=1;i<str.length();i++,cur=cur.next){
cur.next = new Node(str.charAt(i));
}
return head;
}
// Method for printing linked list
public static void printList(Node head){
Node cur = head;
if(head==null)
System.out.print("<EMPTY>");
for(;cur!=null;cur=cur.next){
System.out.print("-> "+cur.item+" ");
}
}
}
答案 0 :(得分：0)
问题与这些代码段中的prev变量的分配有关
public static Node selectionSort_asc(Node head){
...
if(curr.item><min.item) {
prev = min;
min = curr;
}
...
}
public static Node selectionSort_desc(Node head){
...
if(curr.item>max.item) {
prev = max;
max = curr;
}
...
}
您当前正在将prev分配给旧的min和旧的max，但是根据代码的设计，您希望prev指向新min和新max之前的节点，以及从原始链接列表中删除min节点的方式。我的建议是保留另一组名为beforeMin和beforeMax的变量，并在找到新的最小值或最大值时将其设置为等于prev的当前值。除此之外，您还应该像在对prev所做的那样，在while循环的每次迭代中更新curr变量。这是selectionSort_asc的样子，您可以弄清楚如何调整递减方法以反映它：
public static Node selectionSort_asc(Node head){
Node result = null;
Node curr, prev, min, beforeMin;
while(head!=null) {
curr = head;
prev = null;
min = head;
beforeMin = prev; // new variable
while(curr.next!=null) {
prev = curr; // update prev with each iteration
curr = curr.next;
if(curr.item<min.item) {
beforeMin = prev; // variable points to the node before the min node
min = curr;
}
}
//append the min at the end of result list
Node add_min = new Node(min.item);
if(result==null)
result = add_min;
else {
Node last = result;
while(last.next!=null) {
last = last.next;
}
last.next = add_min;
}
//delete the min node form original list
if(min==head) {
head = head.next;
}
else if(min.next==null){
beforeMin.next = null; // no more prev here
}else {
beforeMin.next = beforeMin.next.next; // notice how beforeMin is used now
min.next = null;
}
}
return result;
}