java - 数组中最后一个和第一个字符匹配的最长序列

Question

最近有人问我一个问题，即从字符串数组创建一个序列，以便如果元素 1 的最后一个字符与元素 2 的第一个字符匹配，则字符串元素可以组合。

例如:{"ab", "bc", "cd", "ad", "def", "cedd"} 应返回“abceddef”。

我从上述输入中得到的是“abcdef”。

public class LongestSubstringConsecutiveEnds {

    static StringBuilder sbMax = new StringBuilder();
    static StringBuilder sbTemp;

    public static void main(String[] args) {

        String[] inputStrings = {"ab", "bc", "cd", "ad", "def", "cedd"};
        List<String> inputList = new ArrayList<String>(Arrays.asList(inputStrings));

        for(int i=0; i<inputList.size(); i++) {

            String str = inputList.get(i);
            sbTemp = new StringBuilder(str);
            inputList.remove(str);
            longestSequence(sbTemp, new ArrayList<String>(inputList));
            inputList.add(0, str);
        }

        System.out.println(sbMax.toString());
    }

    private static void longestSequence(StringBuilder tempSubstring, final List<String> inputList) {

        System.out.println(tempSubstring.toString() + inputList);
        if(tempSubstring.length() > sbMax.length()) {
            sbMax.delete(0, sbMax.length());
            sbMax.append(tempSubstring);
        }

        for(int i=0; i<inputList.size(); i++) {

            String inputListString = inputList.get(i);
            char tempStrLastChar = tempSubstring.charAt(tempSubstring.length()-1);
            if(inputListString.charAt(0) == tempStrLastChar) {

                String str = inputList.remove(i);
                longestSequence(tempSubstring.append(inputListString.substring(1)), inputList);
                inputList.add(i, str);
            }
        }
    }
}

Answer 1

根据你的问题:

if the last character of element 1 matches the first character of element 2.

您在问题中描述的部分可以很容易完成:

for (int i = 0; i < strings.length - 1; i++) {
    //   last char of element i         is equal     first char of element i+1                  
    if (strings[i].charAt(strings[i].length()-1) == strings[i+1].charAt(0)) {
        // print element i.
        System.out.print(strings[i]);
    }
}

输出:

cd

即，位置 3 与 4 匹配 (cd-def)

但这不匹配

should return "abceddef" And I can't find a logic... where last ef comes from? you mean match when for example last is a and first is b ??. That would be:

for (int i = 0; i < strings.length - 1; i++) {
    // get last and first char
    String actual = strings[i];
    char last  = actual.charAt(actual.length()-1);
    char first = strings[i+1].charAt(0);

    if ((int) first == last + 1) {
        System.out.print(actual);
    }
}

输出:

ab

即位置 2 与 3 匹配 (ab-cd)