我希望能够创建一个程序，用于识别数字中的特定模式。我不确定这是否需要算法或仅需要精心设计的编程。我不需要任何人提供源代码，只需要一些启发性的想法来指导我正确的方向。
数字将固定为6位数，范围从000000到999999。我想每个数字都会存储在数组的一部分中。然后，我希望测试该数字是否符合某种模式。
例如，假设我使用的3种模式是：
我想我卡住的部分是如何将整数数组的每个部分分配给像A或B这样的值。
我的初步想法是将每个部分分配为单独的字母。例如，如果数组包含1 3 5 4 6 8，则我会创建一个类似于下面的映射：
然后就需要使用第一个模式，
XML模式列表如下：
而我的模式匹配异常类看起来像这样
数字将固定为6位数，范围从000000到999999。我想每个数字都会存储在数组的一部分中。然后，我希望测试该数字是否符合某种模式。
例如，假设我使用的3种模式是：
A A A A A A - would match such examples as 111111 , 222222, 333333 etc where
A B A B A B - would match such examples as 121212 , 454545, 919191 etc
A (A+1) (A+2) B (B+1) (B+2) - would match such examples as 123345, 789123, 456234
我想我卡住的部分是如何将整数数组的每个部分分配给像A或B这样的值。
我的初步想法是将每个部分分配为单独的字母。例如，如果数组包含1 3 5 4 6 8，则我会创建一个类似于下面的映射：
A=1
B=3
C=5
D=4
E=6
F=8
然后就需要使用第一个模式，
AAAAAA
测试一下，例如 if (AAAAAA = ABCDEF)，那么我们就匹配了 AAAAAAA。
如果没有匹配成功，则尝试 (ABABAB = ABCDEF) 等等所有的模式。
在这种情况下，C 分配的值和 F 分配的值一样，就像数字 234874 中的情况。
我不确定这是否能让任何人理解，但我想我可以根据反馈来改进我的问题。
总之，我正在寻求有关如何使程序接受一个六位数并返回匹配的模式的想法。
解决方案
在得到评论后，我找到了一个好办法，以下是我创建的最终解决方案。
package com.doyleisgod.number.pattern.finder;
import java.io.File;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.Iterator;
import java.util.Map;
import javax.xml.parsers.DocumentBuilder;
import javax.xml.parsers.DocumentBuilderFactory;
import org.w3c.dom.Document;
import org.w3c.dom.Element;
import org.w3c.dom.Node;
import org.w3c.dom.NodeList;
public class FindPattern {
private final int[] numberArray; //Array that we will match patterns against.
private final Document patternTree = buildPatternTree(); //patternTree containing all the patterns
private final Map<String, Integer> patternisedNumberMap; //Map used to allocate ints in the array to a letter for pattern analysis
private int depth = 0; //current depth of the pattern tree
// take the int array passed to the constructor and store it in out numberArray variable then build the patternised map
public FindPattern (int[] numberArray){
this.numberArray = numberArray;
this.patternisedNumberMap = createPatternisedNumberMap();
}
//builds a map allocating numbers to letters. map is built from left to right of array and only if the number does not exist in the map does it get added
//with the next available letter. This enforces that the number assigned to A can never be the same as the number assigned to B etc
private Map<String, Integer> createPatternisedNumberMap() {
Map<String, Integer> numberPatternMap = new HashMap<String, Integer>();
ArrayList<String> patternisedListAllocations = new ArrayList<String>();
patternisedListAllocations.add("A");
patternisedListAllocations.add("B");
patternisedListAllocations.add("C");
patternisedListAllocations.add("D");
Iterator<String> patternisedKeyIterator = patternisedListAllocations.iterator();
for (int i = 0; i<numberArray.length; i++){
if (!numberPatternMap.containsValue(numberArray[i])) {
numberPatternMap.put(patternisedKeyIterator.next(), numberArray[i]);
}
}
return numberPatternMap;
}
//Loads an xml file containing all the patterns.
private Document buildPatternTree(){
Document document = null;
try {
File patternsXML = new File("c:\\Users\\echrdoy\\Desktop\\ALGO.xml");
DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
DocumentBuilder db = dbf.newDocumentBuilder();
document = db.parse(patternsXML);
} catch (Exception e){
e.printStackTrace();
System.out.println("Error building tree pattern");
}
return document;
}
//gets the rootnode of the xml pattern list then called the dfsnodesearch method to analyse the pattern against the int array. If a pattern is found a
//patternfound exception is thorwn. if the dfsNodeSearch method returns without the exception thrown then the int array didn't match any pattern
public void patternFinder() {
Node rootnode= patternTree.getFirstChild();
try {
dfsNodeSearch(rootnode);
System.out.println("Pattern not found");
} catch (PatternFoundException p) {
System.out.println(p.getPattern());
}
}
//takes a node of the xml. the node is checked to see if it matches a pattern (this would only be true if we reached the lowest depth so must have
//matched a pattern. if no pattern then analyse the node for an expression. if expression is found then test for a match. the int from the array to be tested
//will be based on the current depth of the pattern tree. as each depth represent an int such as depth 0 (i.e root) represent position 0 in the int array
//depth 1 represents position 1 in the int array etc.
private void dfsNodeSearch (Node node) throws PatternFoundException {
if (node instanceof Element){
Element nodeElement = (Element) node;
String nodeName = nodeElement.getNodeName();
//As this method calls its self for each child node in the pattern tree we need a mechanism to break out when we finally reach the bottom
// of the tree and identify a pattern. For this reason we throw pattern found exception allowing the process to stop and no further patterns.
// to be checked.
if (nodeName.equalsIgnoreCase("pattern")){
throw new PatternFoundException(nodeElement.getTextContent());
} else {
String logic = nodeElement.getAttribute("LOGIC");
String difference = nodeElement.getAttribute("DIFFERENCE");
if (!logic.equalsIgnoreCase("")&&!difference.equalsIgnoreCase("")){
if (matchPattern(nodeName, logic, difference)){
if (node.hasChildNodes()){
depth++;
NodeList childnodes = node.getChildNodes();
for (int i = 0; i<childnodes.getLength(); i++){
dfsNodeSearch(childnodes.item(i));
}
depth--;
}
}
}
}
}
}
//for each node at a current depth a test will be performed against the pattern, logic and difference to identify if we have a match.
private boolean matchPattern(String pattern, String logic, String difference) {
boolean matched = false;
int patternValue = patternisedNumberMap.get(pattern);
if (logic.equalsIgnoreCase("+")){
patternValue += Integer.parseInt(difference);
} else if (logic.equalsIgnoreCase("-")){
patternValue -= Integer.parseInt(difference);
}
if(patternValue == numberArray[depth]){
matched=true;
}
return matched;
}
}
XML模式列表如下：
<?xml version="1.0"?>
<A LOGIC="=" DIFFERENCE="0">
<A LOGIC="=" DIFFERENCE="0">
<A LOGIC="=" DIFFERENCE="0">
<A LOGIC="=" DIFFERENCE="0">
<pattern>(A)(A)(A)(A)</pattern>
</A>
<B LOGIC="=" DIFFERENCE="0">
<pattern>(A)(A)(B)(A)</pattern>
</B>
</A>
<B LOGIC="=" DIFFERENCE="0">
<A LOGIC="=" DIFFERENCE="0">
<pattern>(A)(A)(B)(A)</pattern>
</A>
<B LOGIC="=" DIFFERENCE="0">
<pattern>(A)(A)(B)(B)</pattern>
</B>
</B>
</A>
<A LOGIC="+" DIFFERENCE="2">
<A LOGIC="+" DIFFERENCE="4">
<A LOGIC="+" DIFFERENCE="6">
<pattern>(A)(A+2)(A+4)(A+6)</pattern>
</A>
</A>
</A>
<B LOGIC="=" DIFFERENCE="0">
<A LOGIC="+" DIFFERENCE="1">
<B LOGIC="+" DIFFERENCE="1">
<pattern>(A)(B)(A+1)(B+1)</pattern>
</B>
</A>
<A LOGIC="=" DIFFERENCE="0">
<A LOGIC="=" DIFFERENCE="0">
<pattern>(A)(B)(A)(A)</pattern>
</A>
<B LOGIC="+" DIFFERENCE="1">
<pattern>(A)(B)(A)(B+1)</pattern>
</B>
<B LOGIC="=" DIFFERENCE="0">
<pattern>(A)(B)(A)(B)</pattern>
</B>
</A>
<B LOGIC="=" DIFFERENCE="0">
<A LOGIC="=" DIFFERENCE="0">
<pattern>(A)(B)(B)(A)</pattern>
</A>
<B LOGIC="=" DIFFERENCE="0">
<pattern>(A)(B)(B)(B)</pattern>
</B>
</B>
<A LOGIC="-" DIFFERENCE="1">
<B LOGIC="-" DIFFERENCE="1">
<pattern>(A)(B)(A-1)(B-1)</pattern>
</B>
</A>
</B>
<A LOGIC="+" DIFFERENCE="1">
<A LOGIC="+" DIFFERENCE="2">
<A LOGIC="+" DIFFERENCE="3">
<pattern>(A)(A+1)(A+2)(A+3)</pattern>
</A>
</A>
</A>
<A LOGIC="-" DIFFERENCE="1">
<A LOGIC="-" DIFFERENCE="2">
<A LOGIC="-" DIFFERENCE="3">
<pattern>(A)(A-1)(A-2)(A-3)</pattern>
</A>
</A>
</A>
</A>
而我的模式匹配异常类看起来像这样
package com.doyleisgod.number.pattern.finder;
public class PatternFoundException extends Exception {
private static final long serialVersionUID = 1L;
private final String pattern;
public PatternFoundException(String pattern) {
this.pattern = pattern;
}
public String getPattern() {
return pattern;
}
}
不确定这些内容是否能帮助到有类似问题的人，如果有任何评论或建议，欢迎听取。