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我试图理解人与人之间的关系。但是，当我运行单元测试时，测试会永远运行，它没有得到结果，而且我的 CPU 使用率很高。有人可以看到我的代码有什么问题吗？

字符串关系是字符串的多行输入，格式为"A , B C , D"where Ais parent ofB和Cis parent of D。

这是代码的默认构造函数和字符串格式的输入。我们不需要检查格式是否正确：

public SeeRelations(String relations){
    this.relations = relations;
}

这是从格式化输入中获取字符串的每一行的辅助函数：

//helper function to get each line of the string
private ArrayList<String> lineRelations(){
    int i;
    ArrayList<String> lineRelations = new ArrayList<String>();
    String[] lines = relations.split("\n");
    for(i = 0; i < lines.length; i++){
        lineRelations.add(lines[i]);
    }
    return lineRelations;
}

这是将输入格式化字符串中的所有关系放入数组列表的函数：

//helper function to put each of the relationship in arraylists
private ArrayList<ArrayList<String>> allRelations(){
    int i;
    ArrayList<ArrayList<String>> allRelations = new ArrayList<ArrayList<String>>();
    ArrayList<String> lineRelations = lineRelations();
    for(i = 0; i < lineRelations.size(); i++){
        ArrayList<String> eachLine = new ArrayList<String>(Arrays.asList(lineRelations.get(i).split("\\s*,\\s*")));
        allRelations.add(eachLine);
    }
    return allRelations;
}

这是检查输入名称是否存在的方法：

//helper function to see if the name exist for seeRelations()
private boolean hasThisName(String name){
    ArrayList<ArrayList<String>> allRelations = allRelations();
    int i;
    int j;
    for(i = 0; i < allRelations.size(); i++){
        for(j = 0; j < allRelations.get(i).size(); j++){
            if(name.equals(allRelations.get(i).get(j))){
                return true;
            }
        }
    }
    return false;
}

这是获取两个人之间代号的函数：

//helper function to get Generation number of seeRelations()
private int getGenerationNum(String person, String ancestor){
    ArrayList<ArrayList<String>> allRelations = allRelations();
    String name;
    int i;
    int j;
    int generationNum = 0;
    for(i = 0, j = 0, name = ancestor; i < allRelations.size(); i++){
        if(name.equals(allRelations.get(i).get(0)) && !person.equals(allRelations.get(i).get(1))){
            generationNum++;
            ancestor = allRelations.get(i).get(1);
            i = 0;
            j = 1;
        }
        else if(ancestor.equals(allRelations.get(i).get(0)) && person.equals(allRelations.get(i).get(1))){
            generationNum++;
            j = 1;
            break;
        }
    }
    if(j == 0){
        return 0;
    }
    else{
        return generationNum;
    }
}

"great"这是为最终输出获取倍数的方法：

private String great(int num){
    int i;
    String great = "";
    for(i = 0; i < num; i++){
        great += "great";
    }
    return great;
}

这是我检查两个人之间关系的最后一种方法：

public String SeeRelations(String person, String ancestor){
    int generationNum = getGenerationNum(person, ancestor);
    String great = great(generationNum  - 2);
    if(!(hasThisName(person) && hasThisName(ancestor))){
        return null;
    }
    else{
        if(generationNum == 0){
            return null;
        }
        else if(generationNum == 1){
            return ancestor + " is the parent of " + person;
        }
        else if(generationNum == 2){
            return ancestor + " is the grandparent of " + person;
        }
        else{
            return ancestor + " is the" + " " +  great +"grandparent of " + person;
          }
    }
}

这是我的测试用例，它永远运行，无法得到结果

public class FamilyTreeTest {

    @Test
    public void testSeeRelations() {
        FamilyTree relation2 = new FamilyTree("John Doe ,   Mary Smith" + "\n" + "Martin Weasel ,  John Doe");
        assertEquals("Martin Weasel is the grandparent of Mary Smith", familyTree2.SeeRelations("Mary Smith", "Martin Weasel"));
    }
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1 回答 1

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 for(i = 0, j = 0, name = ancestor; i < allRelations.size(); i++){
            if(name.equals(allRelations.get(i).get(0)) && !person.equals(allRelations.get(i).get(1))){
                generationNum++;
                ancestor = allRelations.get(i).get(1);
                i = 0;
                j = 1;
            }
            else if(ancestor.equals(allRelations.get(i).get(0)) && person.equals(allRelations.get(i).get(1))){
                generationNum++;
                j = 1;
                break;
            }
        }

在这里，您的案例中有错误的行，您的祖先/名字是“Martin Weasel”，因为马丁的关系是“John Doe”，但是您正在寻找玛丽史密斯，所以name.equals(allRelations.get(i).get(0)) && !person.equals(allRelations.get(i).get(1)))这是这样true，这 i = 0;使您的循环从头开始

你可以做什么，尝试创建对象 person 即 class Person{ String name; 列出孩子；列出父母；... } 然后只做简单的tree walker

int SearchDown(Person person, String searchedRelation,int generation)
{
if person.getName().equals(searchedRelation())
return generation;
for (Person child: person.getChildren())
{
int generation = SearchDown(child, searchedRelation, generation+1);
if (generation!=-1) return generation;
}
return -1;
}

ETC...

我真的发现这种方式更容易处理所有类型的树

于 2012-11-19T10:09:20.987 回答