java ancestor_java – 家谱祖先查找算法
我的方法無法通過單元測試.我盯著它看了5個小時是徒勞的.有人可以幫我看看它有什么問題嗎?
PS:我的代碼中的getAllRelations()方法是將格式化的輸入分離為字符串ArrayList的arraylists,例如,如果我有這樣的格式化輸入(我在我的測試用例中使用它無法通過)
String format = "John Doe , Mary Smith" + "\n" + "Brian William , John Doe" + "\n" + "Brian William ,Robert Andrew" + "\n" + "Mary Smith , Max Jackson";
在每一行中,第一個人是第二個人的父母.
getAllRelations()方法將這個格式化的字符串拆分為arraylists,每個列表只包含每行中的兩個名稱字符串(在名稱之前或之后沒有空格)作為其元素.例如arraylist1將是一個包含“John”和“Mary Smith”的列表.
這是我的方法,我無法弄清楚什么是錯的,我想用這種方法來檢查兩個人是否共享同一個祖先.
private boolean hasSameAncestor(String person1, String person2){
ArrayList> allRelations = allRelations();
int i = 0;
int j = 0;
String name1 = person1;
String name2 = person2;
String parent1;
String parent2;
for(i = 0, parent1 = ""; i < allRelations.size(); i++){
if(name1.equals(allRelations.get(i).get(1))){
parent1 = allRelations.get(i).get(0);
for(j = 0, name2 = person2, parent2 = ""; j < allRelations.size(); j++){
if(name2.equals(allRelations.get(j).get(1))){
parent2 = allRelations.get(j).get(0);
if(parent2.equals(parent1)){
return true;
}
else{
name2 = parent2;
j = 0;
}
}
}
name1 = parent1;
i = 0;
}
}
return false;
}
我無法通過的測試用例就是這樣.
@Test
public void testHasSameAncestor()
FamilyTree familyTree4 = new FamilyTree("John Doe , Mary Smith" + "\n" + "Brian William , John Doe" + "\n" + "Brian William ,Robert Andrew" + "\n" + "Mary Smith , Max Jackson");
assertEquals(true, format.hasSameAncestor("Max Jackson", "Robert Andrew"));
}
我無法弄清楚我的功能有什么問題,有人可以幫幫我嗎?非常感謝你.
可以粘貼到eclipse以獲得調試幫助的代碼
package test;
import java.util.ArrayList;
import java.util.Arrays;
public class Test1 {
String test;
public Test1(String test){
this.test = test;
}
private ArrayList lineRelations(){
int i;
ArrayList lineRelations = new ArrayList();
String[] lines = test.split("\n");
for(i = 0; i < lines.length; i++){
lineRelations.add(lines[i]);
}
return lineRelations;
}
private ArrayList> allRelations(){
int i;
ArrayList> allRelations = new ArrayList>();
ArrayList lineRelations = lineRelations();
for(i = 0; i < lineRelations.size(); i++){
ArrayList eachLine = new ArrayList(Arrays.asList(lineRelations.get(i).split("\\s*,\\s*")));
allRelations.add(eachLine);
}
return allRelations;
}
public boolean hasSameAncestor(String person1, String person2){
ArrayList> allRelations = allRelations();
int i = 0;
int j = 0;
String name1 = person1;
String name2 = person2;
String parent1;
String parent2;
for(i = 0, parent1 = ""; i < allRelations.size(); i++){
if(name1.equals(allRelations.get(i).get(1))){
parent1 = allRelations.get(i).get(0);
for(j = 0, name2 = person2, parent2 = ""; j < allRelations.size(); j++){
if(name2.equals(allRelations.get(j).get(1))){
parent2 = allRelations.get(j).get(0);
if(parent2.equals(parent1)){
return true;
}
else{
name2 = parent2;
j = 0;
}
}
}
name1 = parent1;
i = 0;
}
}
return false;
}
}
測試用例
package test;
import static org.junit.Assert.*;
import test.Test1;
import org.junit.Test;
public class Test1Test {
@Test
public void testHasSameAncestor(){
Test1 test1 = new Test1("John Doe , Mary Smith" + "\n" + "Brian William , John Doe" + "\n" + "Brian William ,Robert Andrew" + "\n" + "Mary Smith , Max Jackson");
assertEquals(true, test1.hasSameAncestor("Max Jackson", "Robert Andrew"));
}
}
《新程序員》:云原生和全面數字化實踐50位技術專家共同創作,文字、視頻、音頻交互閱讀總結
以上是生活随笔為你收集整理的java ancestor_java – 家谱祖先查找算法的全部內容,希望文章能夠幫你解決所遇到的問題。
- 上一篇: java exe 返回值_java调用e
- 下一篇: java finalize逃脱_关于Ja