我有一个 vector 数组列表,我试图找到所有可能的不同路径(不重复相同的节点)。但是保存路径记录的 ArrayList 似乎是共享的,因此会在所有路径中泄漏。我的问题是,是否有我应该使用的替代列表类型来防止这种情况发生?谢谢。
可用路径为(A->B、B->C、C->D、D->A、A->C)
首先,我决定只处理起始节点 A 的路径。
输出应该是:
ABCD
ACD
我编写了下面的代码,并附加了输出来检查发生了什么,因为它无法正常工作。下面代码的输出是:
上一个 vector :SA
vector 历史大小:1
下一个 vector :AB
上一个 vector :AB
vector 历史大小:2
下一个 vector :BC
上一个 vector :BC
vector 历史大小:3
下一个 vector :CD
上一个 vector :CD
vector 历史大小:4
下一个 vector :DA
循环 - ABCDA
ABCDA
ABC
AB
vector 历史大小:4
下一个 vector :AC
循环 - AC
交流
正如您所看到的,while 循环的第二次迭代的最后 4 行是错误的,因为 vector 历史大小应该为 1(仅限 SA)并且“C”之前不应该被访问过,但不知何故 vector 历史的 ArrayList从第一个 while 循环的递归开始就已经泄漏了。这种情况是否会发生以及有哪些替代方案?
public static void main(String[] args) {
ArrayList<vector> vectorList = new ArrayList();
vectorList.add(new vector("A", "B"));
vectorList.add(new vector("B", "C"));
vectorList.add(new vector("C", "D"));
vectorList.add(new vector("D", "A"));
vectorList.add(new vector("A", "C"));
//to record vector history and initialize the start vector
ArrayList<vector> vectorHistory = new ArrayList();
//record the path
String path = "";
//method call
pathFinder(new vector("S", "A"), vectorHistory, vectorList, path);
}
//Recursive method. moves one node forward until there is no more nodes OR the next node is the same as a previously taken node
public static void pathFinder(vector prevVector, ArrayList<vector> vectorHistory, ArrayList<vector> vectorList, String path) {
vectorHistory.add(prevVector);
//add the current node to the path
path = path + prevVector.child;
System.out.println("Previous vector: "+ prevVector.parent+prevVector.child);
// search if there is a next node. looped to search all possible paths
while (vectorList.contains(prevVector)) {
System.out.println("vector history size: "+ vectorHistory.size());
//retrieve the next vector
vector nextVector = vectorList.get(vectorList.indexOf(prevVector));
System.out.println("Next vector: " + nextVector.parent + nextVector.child);
//remove current node so while loop can move to another possible path
vectorList.remove(vectorList.indexOf(prevVector));
//check if the next node has already been visited before
if (vectorHistory.contains(nextVector)) {
path=path+nextVector.child;
System.out.println("Looped - " + path);
} else {
pathFinder(nextVector, vectorHistory, vectorList, path);
}
}
System.out.println(path);
}
/*object vector */
public static class vector {
String parent, child;
public vector(String parent, String child) {
this.parent = parent;
this.child = child;
}
@Override
public boolean equals(Object o) {
vector x = (vector) o;
if (x.parent.equalsIgnoreCase(child)) {
return true;
} else {
return false;
}
}
}
最佳答案
Java 是“按值传递”的,因此它传递实际对象的引用的副本。但是使用集合时理解起来有点奇怪,因为发送的引用的副本指向与原始引用相同的内存!
因此,如果您将列表传递给方法并修改列表中的方法,它将修改原始列表。
例如:
method b(List aList){
aList.add(new Object());
}
method c(List aList){
aList=new ArrayList ();
aList.add(new Object());
}
List a=new ArrayList();
b(a); -> it will add an object to a;
c(a); -> it will not add an object to a or modify it in any way
所以就你的情况来说,当你打电话时
pathFinder(nextVector, vectorHistory, vectorList, path);
您不会获得您期望的递归“堆栈”行为,因为路径查找器的后继调用会修改前一个列表。
您可以这样修改您的调用:
pathFinder(nextVector, new ArrayList<>(vectorHistory), new ArrayList<>(vectorList), path);
为了避免这个问题,但每次复制整个列表都会丢失一些额外的内存;)并且它仍然不会得到你想要的结果,因为我猜你的算法中有另一个错误。
你的程序看起来很奇怪;)你用 vector 的相等所做的魔法并不大,因为你实际上无法比较两个相等的对象。例如,您的代码 AB 与 AB 不同(事实并非如此)。因此,对于你去过的地方,你不需要 vector ,而是点。所以这里有一个稍微修改过的程序只是为了说明我的意思。它还远未达到完美:
import java.util.ArrayList;
import java.util.List;
public class MainClass {
public static void main(String[] args) {
List<MyVector> vectorList = new ArrayList<MyVector>();
vectorList.add(new MyVector("A", "B"));
vectorList.add(new MyVector("B", "C"));
vectorList.add(new MyVector("C", "D"));
vectorList.add(new MyVector("D", "A"));
vectorList.add(new MyVector("A", "C"));
List<String> pointsHistory=new ArrayList<String>();
//to record points that have been visited
//record the path
String path = "";
//method call
pathFinder(new MyVector(null, "A"), pointsHistory, vectorList, path);
}
//Recursive method. moves one node forward until there is no more nodes OR the next node is the same as a previously taken node
public static void pathFinder(MyVector prevVector, List<String> pointsHistory, List<MyVector> vectorList, String path) {
pointsHistory.add(prevVector.child);
//add the current node to the path
path = path + prevVector.child;
// search if there is a next node. looped to search all possible paths -> no need to do magic with equals
for(MyVector vector:vectorList)
if(vector.parent.equals(prevVector.child)) {
System.out.println("Next vector: " + vector.parent + vector.child);
if (pointsHistory.contains(vector.child)) {
System.out.println("Result " + path); //You get the end result here -> if we have reached a loop
} else {
pointsHistory.add(vector.child);
pathFinder(vector, new ArrayList<>(pointsHistory), vectorList, path);
}
}
}
/*object vector */
public static class MyVector {
String parent, child;
public MyVector(String parent, String child) {
this.parent = parent;
this.child = child;
}
}
}
这样你就会得到你想要的结果。看看我如何复制访问过的点:pathFinder(vector, new ArrayList<>(pointsHistory), vectorList, path);为了这项工作。并请用大写字母命名您的类(class)。
关于java - 递归方法中 ArrayList 的行为,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/50022791/