Java -- idiom chain -- recursive algorithm -- spring JDBC + c3p0

TOTAL 2020-11-08 16:58:51
java idiom chain recursive algorithm


 
This example , I practice SpringJdbc+c3p0 When I wrote . Input an idiom , Get an incomplete elderberry . Because I have... In my database 3W Many idioms , In the case of excluding the repetition of the samurai idiom , Still can't get all the results , I don't think recursion goes into an endless loop , because My recursion has an end condition , When a given idiom , There's no more solitaire , Just go back to . Take a look at the code .
 
This example , Let me know that sometimes , With the code to solve the problem, baidu may have no specific way to try . And then I meet c3p0 When you don't find the right driver , It started with Baidu , It hasn't been solved for a long time , I just follow the code , Find out Again driverManger That's to pass String Of url, But my message is empty ,c3p0 It's also written in the configuration file of , It's all right ,c3p0 Profile name for I copied it on the Internet , Looks like the name of the request , But when I changed the name of the configuration file , I can print a connection . This problem has been solved .
 
after , When I write to find idioms recursively , I've got another question , When I recurs from the root node to the tail node , That is to say, after finding a chain of idioms , I started thinking , How to go back to the next node of the root node , When I do Loop() In the use return Loop() When , You can only get one record . Later, after my rethinking , I just wanted to , When you finish a solitaire , I've copied this solitaire , And remove the value at the end , I don't know how to express it here , I'll stick the code and say .
private List<List<String>>Loop(String word,List<String>listOne,List<List<String>>bigList,int m ){
if(listOne==null){
listOne=new ArrayList<String>();
}
if(bigList==null){
bigList=new ArrayList<>();
}
if(word.equals("")||word==null){
return bigList;
}
Object []args=new Object[1];
args[0]=word.substring(word.length()-1);
List<ChengYu>words=myDb.QueryForBean(selSql,args);
if(words==null||words.isEmpty()){
bigList.add(listOne);
return bigList;
}
for (int i=0;i<words.size();i++) {
ChengYu cy = words.get(i);
Boolean flag=false;
for (List<String> ll:bigList){
if(ll.contains(cy.getWord())){
flag=true;
break;
}
}
if (listOne != null && !listOne.contains(cy.getWord())&&!flag)
{ listOne.add(cy.getWord());}
else{
continue;
}
// Because there are so many situations , So when it exceeds 3 Just go back to
if(m==3){
return bigList;
}
Loop(cy.getWord(), listOne, bigList,m++);
ArrayList copy2 =new ArrayList<String>();
copy2.addAll(listOne);
listOne=null;
listOne=copy2;
listOne.remove(listOne.size()-1);
}
return bigList;
}
Every listOne For a solitaire ,bigList Represents the Solitaire you've got .words It means that there are many idioms under the current idiom , If this words Is empty or null On behalf of listOne It's a complete solitaire , This is the time , Just put listOne Add to bigList, After that return Code return after , Just walk to ArrayList copy2 =new ArrayList<String>(); This sentence , This recursion does not recursion once, it packs the variables in the function and puts them on the stack as a whole ,return Just take one from the stack LooP, therefore , After finishing another idiom , I can get the second idiom from the penultimate Jielong group B, I need idioms from B Continue to recursive , At this time, I need to create a new idiom company , And put B The previous idiom node is added in advance , And then there was the following
ArrayList copy2 =new ArrayList<String>();
 
Because you get B after , And from B Start recursion , I do this by putting B Previous lujing Keep it .
That's what I mean
a1->b1->c1
a1->b1->c2
a1->b2->d3
 
a3->q3->m4->e5
If a The tag is the idiom group following the root node , Then under the root node there is a1 and a3 Two idioms can connect a dragon .
Empathy a1 Yes b1,b2 Two idioms can connect a dragon ,b1 Yes c1,c2...
When listOne(a1,b1,c1) after ,Loop Out of the stack , This is the time to execute
copy2.addAll(listOne);
listOne=null;
listOne=copy2;
listOne.remove(listOne.size()-1);
You can get listOne(a1,b1),for The cycle is to stay and then c This group level loops ,for Loop to the beginning , This is the time
It becomes listOne(a1,b1,c2); If you go on, there will be no child nodes ,Loop Out of the stack ,listOne(a1,b1),, This is the time for stay b layer ,Loop(b1) It's over , Keep going out of the stack , Just keep copying ,listOne It becomes listOne(a1),for b Back to where the loop started ,listOne(a1,b2) , Begin Loop(b2),b2 The following idiom groups are d3,for loop d3 On that floor ,listOne(a1,b2,d3),Loop(d3) because d3 No successor nodes ,Loop(d3) Out of the stack ,listOne(a1,b2), here Loop(b2) All call stacks belong to end , Loop(b2) Out of the stack ,listOne(a1), This is the time Loop(a1) End of call stack for ,Loop(a1) Out of the stack ,listOne(),for The cycle stops at a layer ,list(a3),Loop(a3),list(a3,q3),Loop(q3),list(a3,q3,m4),Loop(m4),list(a3,q3,m4,e5),Loop(e5),words Collection is empty Loop(e5) Out of the stack ,listOne(a3,q3,m4).........
Finally back to Loop(a3) Out of the stack ,listOne();
The call stack is empty ,return bigList ; At this point, the whole process is completed .
 
I also expressed recursion for the first time The call stack , It's definitely not clear , If you want to know about recursion, you can search the function call stack on the Internet, and you may understand it .


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The code and sql Script json File download

Extraction code :801w

 
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