Question

所以我在Stack Overflow上看到，可以使用不使用递归的方法派生通用后缀。我很好奇能否通过递归实现相同的目标。尝试使用下面的代码，但得到java.lang.StackOverflowError。有人可以帮忙吗？

public static String commonSuffixWithRecursion(String s1, String s2) {
    if(s1.length()==0 || s2.length()==0) {
        return "Strings are empty";
    }
    if(s1.charAt(s1.length()-1)==s2.charAt(s2.length()-1)) {
        return s1.length()-1 + commonSuffixWithRecursion(s1,s2);
    }
    return "No common Suffix";
}

Answer 1

没有通用后缀时，应返回空字符串。
您可以使用subString获取下一轮的输入
当前字符应在下一次漫游结果后的之后后添加

示例代码：

public static String commonSuffixWithRecursion(String s1, String s2) {
    if (s1.length() == 0 || s2.length() == 0) {
        return "";
    }
    if (s1.charAt(s1.length() - 1) == s2.charAt(s2.length() - 1)) {
        String nextS1 = s1.substring(0, s1.length() - 1);
        String nextS2 = s2.substring(0, s2.length() - 1);
        return commonSuffixWithRecursion(nextS1, nextS2 ) + s1.charAt(s1.length() - 1);
    }
    return "";
}

测试用例：

public static void main(String[] args) {
    System.out.println(commonSuffixWithRecursion("dbc", "abc"));    // bc
    System.out.println(commonSuffixWithRecursion("a", "a"));    // a
    System.out.println(commonSuffixWithRecursion("ab", "b"));   // b
    System.out.println(commonSuffixWithRecursion("a", "b"));    // empty
}

Answer 2

一种简单的递归方法是

public static String commonSuffixWithRecursion(String x, String y, int m, int n) {
        if(m <= 0 || n <= 0)
            return "";
        if (x.charAt(m-1) == y.charAt(n-1)) 
            return commonSuffixWithRecursion(x, y, m-1, n-1) + x.charAt(m-1);
        else 
         return "";
    }

这里m=x.length() and n=y.length()

Answer 3

这是后缀大小增加的另一种选择：

public static String commonSuffixWithRecursion(String s1, String s2, int suffixSize) {
    int i1 = s1.length() - suffixSize;
    int i2 = s2.length() - suffixSize;
    if (i1 > -1 && i2 > -1 && s1.charAt(i1) == s2.charAt(i2)) {
        return commonSuffixWithRecursion(s1, s2, suffixSize + 1);
    }
    return s1.substring(i1 + 1);
}

以下是一些测试用例：

public static void main(String[] args) {
    System.out.println(commonSuffixWithRecursion("dbc", "abc", 1));    // bc
    System.out.println(commonSuffixWithRecursion("a", "a", 1));    // a
    System.out.println(commonSuffixWithRecursion("ab", "b", 1));   // b
    System.out.println(commonSuffixWithRecursion("a", "b", 1));    // empty
    System.out.println(commonSuffixWithRecursion("a", "bc", 1));    // empty
    System.out.println(commonSuffixWithRecursion("ac", "bc", 1));    // c
    System.out.println(commonSuffixWithRecursion("ac", "bcd", 1));    // empty
    System.out.println(commonSuffixWithRecursion("", "bcd", 1));    // empty
    System.out.println(commonSuffixWithRecursion("bcd", "bcd", 1));    // bcd
}

如何在Java中使用递归查找常见的后缀

3 个答案: