Java程序故障

Question

我的问题的前半部分：当我尝试运行我的程序时，它会永远加载并加载；它从不显示结果。有人可以检查我的代码并在某处发现错误。该程序旨在找到一个起始 DNA 密码子 ATG 并不断寻找，直到找到一个终止密码子 TAA 或 TAG 或 TGA，然后从头到尾打印出基因。我正在使用 BlueJ。

我的问题的后半部分：我应该编写一个需要执行以下步骤的程序：

To find the first gene, find the start codon ATG.
Next look immediately past ATG for the first occurrence of each of the three stop codons TAG, TGA, and TAA.
If the length of the substring between ATG and any of these three stop codons is a multiple of three, then a candidate for a gene is the start codon through the end of the stop codon.
If there is more than one valid candidate, the smallest such string is the gene. The gene includes the start and stop codon.
If no start codon was found, then you are done.
If a start codon was found, but no gene was found, then start searching for another gene via the next occurrence of a start codon starting immediately after the start codon that didn't yield a gene.
If a gene was found, then start searching for the next gene immediately after this found gene.

请注意，根据此算法，对于字符串“ATGCTGACCTGATAG”，ATGCTGACCTGATAG 可能是一个基因，但 ATGCTGACCTGA 不是，即使它更短，因为首先找到另一个不是倍数的 'TGA' 实例三个远离起始密码子。

在我的作业中，我也被要求提供这些方法：

具体而言，要实现该算法，您应该执行以下操作。

Write the method findStopIndex that has two parameters dna and index, where dna is a String of DNA and index is a position in the string. This method finds the first occurrence of each stop codon to the right of index. From those stop codons that are a multiple of three from index, it returns the smallest index position. It should return -1 if no stop codon was found and there is no such position. This method was discussed in one of the videos.
Write the void method printAll that has one parameter dna, a String of DNA. This method should print all the genes it finds in DNA. This method should repeatedly look for a gene, and if it finds one, print it and then look for another gene. This method should call findStopIndex. This method was also discussed in one of the videos.
Write the void method testFinder that will use the two small DNA example strings shown below. For each string, it should print the string, and then print the genes found in the string. Here is sample output that includes the two DNA strings:

示例输出为：

ATGAAAATGAAAA

找到的基因是：

ATGAAAATGA

DNA 字符串是：

ccatgccctaataaatgtctgtaatgtaga

发现的基因是：

atgccctaa

atgtctgtaatgtag

DNA 字符串是：

CATGTAATAGATGAATGACTGATAGATATGCTTGTATGCTATGAAAATGTGAAATGACCCA

发现的基因是：

ATGTAA

ATGAATGACTGATAG

ATGCTATGA

ATGTGA

我已经考虑过了，发现这段代码接近正常工作状态。我只需要让我的输出产生说明中要求的结果。希望这不会太混乱，我只是不知道如何在起始密码子之后寻找终止密码子，然后如何获取基因序列。我也希望了解如何通过查找三个标签（tag、tga、taa）中的哪个更接近 atg 来获得最接近的基因序列。我知道这很多，但希望这一切都有意义。

import edu.duke.*;
import java.io.*;

public class FindMultiGenes {
    public String findGenes(String dnaOri) {
        String gene = new String();
        String dna = dnaOri.toLowerCase();
        int start = -1;
        while(true){
            start = dna.indexOf("atg", start);
            if (start == -1) {
                break;
            }
            int stop = findStopCodon(dna, start); 
            if(stop > start){
                String currGene = dnaOri.substring(start, stop+3);

                System.out.println("From: " + start + " to " + stop + "Gene: "    
                +currGene);}
        }
        return gene;
    } 

    private int findStopCodon(String dna, int start){   
        for(int i = start + 3; i<dna.length()-3; i += 3){
            String currFrameString = dna.substring(i, i+3);

            if(currFrameString.equals("TAG")){
                return i;

            } else if( currFrameString.equals("TGA")){
                return i;

            } else if( currFrameString.equals("TAA")){
                return i;

            }
        }   
        return -1;
    }

    public void testing(){


        FindMultiGenes FMG = new FindMultiGenes();

        String dna =     
        "CATGTAATAGATGAATGACTGATAGATATGCTTGTATGCTATGAAAATGTGAAATGACCCA";

        FMG.findGenes(dna);




        System.out.println("DNA string is: " + dna);

    } 
}   

Answer 1

把你的start = dna.indexOf("atg", start);改成

start = dna.indexOf("atg", start + 1);

当前发生的情况是您在索引k 处找到"atg"，并在下一次运行中搜索字符串以查找从k 开始的下一个"atg"。这会在完全相同的位置找到下一个匹配项，因为起始位置包含在内。因此，您将一遍又一遍地找到相同的索引k，并且永远不会停止。

通过将索引增加 1，您可以跳过当前找到的索引 k 并开始从 k+1 开始搜索下一个匹配项。

Answer 2

这个程序是为了找到一个起始 DNA 密码子 ATG 并不断寻找直到找到一个终止密码子 TAA 或 TAG 或 TGA，然后从头到尾打印出基因。

由于第一次搜索总是从 0 开始，您可以在此处设置起始索引，然后从结果中搜索终止密码子。在这里，我使用 1 个终止密码子：

public static void main(String[] args) {

    String dna = "CATGTAATAGATGAATGACTGATAGATATGCTTGTATGCTATGAAAATGTGAAATGACCCA";
    String sequence = dna.toLowerCase();
    int index = 0;
    int newIndex = 0;
    while (true) {
        index = sequence.indexOf("atg", index);
        if (index == -1)
            return;
        newIndex = sequence.indexOf("tag", index + 3);
        if (newIndex == -1) // Check needed only if a stop codon is not guaranteed for each start codon.
            return;
        System.out.println("From " + (index + 3) + " to " + newIndex + " Gene: " + sequence.substring(index + 3, newIndex));
        index = newIndex + 3;
    }
}

输出：

From 4 to 7 Gene: taa
From 13 to 22 Gene: aatgactga

此外，您可以使用正则表达式为您完成大量工作：

public static void main(String[] args) {

    String dna = "CATGTAATAGATGAATGACTGATAGATATGCTTGTATGCTATGAAAATGTGAAATGACCCA";

    Pattern p = Pattern.compile("ATG([ATGC]+?)TAG");
    Matcher m = p.matcher(dna);

    while (m.find())
        System.out.println("From " + m.start(1) + " to " + m.end(1) + " Gene: " + m.group(1));
}

输出：

From 4 to 7 Gene: TAA
From 13 to 22 Gene: AATGACTGA