我希望创建一个程序来识别数字中的某些模式。我不确定这是否需要算法或只是经过深思熟虑的编程。我不是在寻找提供源代码的人，只是在寻找一些发人深省的想法，让我朝着正确的方向前进。
数字将固定长度为 6 位数字，从 000000 到 999999。我猜每个数字都将存储为数组的一部分。然后我想根据模式测试数字。
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一个值，例如 A 或 B
我最初的想法只是将每个部分分配为一个单独的字母。因此，如果数组由 1 3 5 4 6 8 组成，那么我将创建一个像
这样的 mapA=1
B=3
C=5
D=4
E=6
F=8
然后一些如何采取第一个模式，
AAAAAA
并用类似 if (AAAAAA = ABCDEF) 的东西进行测试，然后我们匹配了 AAAAAAA
如果不是，则尝试 (ABABAB = ABCDEF) 等遍历我的所有模式
在这种情况下，分配给 C 的值没有理由不能与分配给 F 的值相同，例如数字 234874。
我不确定这是否对任何人都有意义，但我想我可以根据反馈完善我的问题。
总而言之，我正在寻找有关如何让程序接受 6 位数字并返回给我们它匹配的模式的想法。
解决方案
在给出的评论让我走上正轨之后，下面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;
}
}
模式的 xml 列表如下所示
<?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>
我的模式发现异常类如下所示
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;
}
}
不确定这是否会对遇到类似问题的任何人有所帮助，或者如果有人对它的工作方式有任何意见，我会很高兴听到他们的意见。
最佳答案
我建议构建状态机:
一个。使用模式初始化:
- 规范化所有模式
- 构建一棵深度为 6 的树，表示从根开始的所有模式以及每个深度的所有可能选择。
B.运行状态机
A.1。模式归一化。
A A A A A A => A0 A0 A0 A0 A0 A0
C A C A C A => A0 B0 A0 B0 A0 B0(始终以 A 开头，然后是 B、C、D 等)
B B+1 B+2 A A+1 A+2 => A0 A1 A2 B0 B1 B2
因此，您始终拥有以 A0 开头的规范化模式。
A.2。建一棵树
1. A0
/ | \
2. A0 B0 A1
| | |
3. A0 A0 A2
| | |
4. A0 B0 B0
| | |
5. A0 A0 B1
| | |
6. A0 B0 B2
| | |
p1 p2 p3
B.运行状态机
使用Depth-first search algorithm使用递归查找匹配的模式。
有意义吗？
关于用于识别数字模式的 Java 程序，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/11422498/