获取词典上最小和最大的子字符串。我的算法失败了大多数测试案例，但我不明白为什么。帮我弄明白

Question

我今天在面试中做了一个测试，问题是获取词典上最小和最大的子字符串（换句话说，按名称排序）。

link-complete函数SmallestAndLargestSubstring，该函数将由小写英文字母（a-z）组成的字符串S作为参数，并返回以元音开头、以辅音结尾的最小和最大的子字符串。

我的算法通过了基本的测试用例，但没有通过其他的大多数测试用例。这不是最有效的代码，但它是编写速度最快的。

static String[] SmallestAndLargestSubstring(String s) {

    ArrayList<Character> vowelList = new ArrayList<Character>();
    vowelList.add('a');
    vowelList.add('e');
    vowelList.add('i');
    vowelList.add('o');
    vowelList.add('u');

    ArrayList<Character> consonantList = new ArrayList<Character>();
    for (char c='a'; c<='z'; c++) {
        if (!vowelList.contains(c))
            consonantList.add(c);
    }

    ArrayList<String> substringList = new ArrayList<String>();

    for (int i=0; i<s.length(); i++) {
        char c = s.charAt(i);
        if (vowelList.contains(c)) {
            String substring = "";
            substring+=c;
            for (int j=i+1; j<s.length(); j++) {
                char c2 = s.charAt(j);
                substring+=c2;
                if (consonantList.contains(c2)) {
                    substringList.add(substring);
                }
            }
        }
    }

    Collections.sort(substringList);

    String[] outputAdapter = new String[2];
    outputAdapter[0]=substringList.get(0);
    outputAdapter[1]=substringList.get(substringList.size()-1);

    return outputAdapter;
}

输入

字符串s=“Azizezozuzawwwwwwwwuzzzzzzzzzbbbbbbbbaaaaabbbbbboiz”

我的回答

编辑：这里还有3个测试用例。我的答案与这些测试用例的答案相匹配。

string=“ABA”；最小=“AB”；最大=“AB”；

string=“AAB”；最小=“AAB”；最大=“AB”；

string=“abababaaaaaaaaaaaaaaaaaaaaaaaaaaz”；最小=“AAAAAAAAAAAAAAAAAAAZ”；最大=“AZ”；

Answer 1

/*
It is the Basic code to Obtain Substring which start with Vowel and End up with Consonant. It is going to print on the Basis of Name Comparable, The first and the Last Substring in the List.Similarly we can achieve on the basis of length, the firt and last Substring using different comparator function.
*/

public class StringSubsequencesStartVowelEndConsonant {
    static List<String> subsequence = new LinkedList<>();

    public static void main(String[] args) {        
        Scanner in = new Scanner(System.in);
        System.out.println("Enter the String:\n");
        String string = in.next();
        findSubstring(string);      
    }

    private static void findSubstring(String string) {      
        for(int i=0;i<string.length();i++){
            if(isVowel(string.charAt(i))){              
                for(int j=string.length()-1;j>=i;j--){
                    if(isConsonant(string.charAt(j))){                      
                        String subString = string.substring(i,j+1);
                        subsequence.add(subString);
                    }
                }
            }
        }       
        Collections.sort(subsequence);
        for(String str : subsequence){
            System.out.print(str+" ");
        }
        System.out.println();
        System.out.println(subsequence.get(0));
        System.out.println(subsequence.get(subsequence.size()-1));      
    }
    private static boolean isConsonant(char chars) {
        return !(chars=='a'|| chars=='e'||chars=='i'||chars=='o'||chars=='u');
    }
    private static boolean isVowel(char chars) {
        return (chars=='a'|| chars=='e'||chars=='i'||chars=='o'||chars=='u');
    }
}