java - 关于单词搜索算法的问题

Question

我的代码似乎抛出此错误:java.lang.StringIndexOutOfBoundsException:字符串索引超出范围:0 使用 substring 和 string.charAt() 方法

我尝试更改 for 循环和参数中的值，因此字符串索引不会超过其预期值

public final class WordSearcher {
    public static void main(String args[]) {
        //String wordLibrary = args[0];
        //String puzzle = args[1];
        int j,k,l;
        String wordLibrary = "TEST,SAMPLE,OUTPUT";
        String puzzle = "TUPTUOELPMAS";
        String puzzleDup = puzzle;
        String[] wordstoFind = wordLibrary.split(",");
        int ifWordExists = 0;
        int puzzleLength = puzzle.length();

        for(k = 0; k< wordstoFind.length; k++) {
            for (l = 0; l< wordstoFind[k].length(); l++) {
                for(j=0; j< puzzleLength; j++) {

                    if(wordstoFind[k].charAt(l) == puzzleDup.charAt(j)) {
                        ifWordExists++;
                        puzzleDup = puzzleDup.substring(j+1,puzzleDup.length());
                        break;
                    }
                }   
            } 

            if( ifWordExists == wordstoFind[k].length()) 
                    System.out.println(wordstoFind[k]);


                ifWordExists = 0;
                System.out.println(puzzle);
                puzzleDup = puzzle;
        }

            for(k = 0; k< wordstoFind.length; k++) {
                for (l = 0; l< wordstoFind[k].length(); l++) {
                    for(j=puzzleLength-1; j> 0; j--) {

                        if(wordstoFind[k].charAt(l) == puzzleDup.charAt(j)) {
                            ifWordExists++;
                            puzzleDup = puzzleDup.substring(0,j-1);
                            break;
                        }
                    }     
                } 

                if( ifWordExists == wordstoFind[k].length()) 
                    System.out.println(wordstoFind[k]);


                ifWordExists = 0;
                System.out.println(puzzle);
                puzzleDup = puzzle;
        }
    }
}

它应该找到分散在作为命令行参数传递的字符串中的单词。例如，XTXXXEXXXXXSXT 应该返回 TEST，如果这是我必须找到的单词之一。

Answer 1

这里:

for(k = 0; k < wordstoFind.length; k++) {
    for (l = 0; l < wordstoFind[k].length(); l++) {
        for(j = 0; j < puzzleLength; j++) {
            if(wordstoFind[k].charAt(l) == puzzleDup.charAt(j)) {
                ifWordExists++;
                puzzleDup = puzzleDup.substring(j + 1, puzzleDup.length());
                                                ^^^^^
                break;
            }
        }   
    }

如果j等于puzzleLength - 1，j + 1将超出字符串的范围。