最长的常见子序列java(递归)

时间:2016-11-22 19:45:08

标签: java longest-substring recursive-backtracking

我正在处理的问题是: http://practiceit.cs.washington.edu/problem/view/cs2/sections/recursivebacktracking/longestCommonSubsequence

基本上我们有两个字符串,我们被要求找到最长的公共子序列。我已经在网上搜索了解决方案,并将它们与我自己的解决方案进行了比较,我在代码中找不到任何错误。我想知道为什么它仍然无法运作。

而且,我被要求通过使用递归方法来解决这个问题

这是我的代码:

public static String longestCommonSubsequence(String a, String b){
    if(a.isEmpty() || b.isEmpty()){
        return "";
    }
    if (a.substring(a.length() - 1).equals(b.substring(b.length() - 1))){
        return longestCommonSubsequence(a.substring(0, a.length() - 1), b.substring(0, b.length()
                       - 1)) + a.substring(a.length() - 1);
    } else {
        String first = longestCommonSubsequence(a, b.substring(b.length() - 1));
        String second = longestCommonSubsequence(a.substring(a.length() - 1), b);
        if(first.length() > second.length()){
            return first;
        }
        return second;
    }
}

以下是所有测试用例:

回复致电价值

" ABCDEFG"," BGCEHAF" " BCEF"

"她出售","贝壳" " sesells"

" 12345"," 54321 21 54321" " 123"

"高傲的老师","美味的桃子" " ecious each"

" Marty"," Helene" ""

"""乔" ""

" Suzy","" ""

" ACGGTGTCGTGCTA"," CGTTCGGCTATCGTACGT" " CGGTTCGTGT"

使用我的代码我获得了所有测试用例的StackOverFlow。

2 个答案:

答案 0 :(得分:0)

您的LCS计算不正确。在LCS中,您需要从字符串的结尾进行比较。如果两个字符串的最后一个字符匹配则意味着它将成为您LCS的一部分。

public static String longestCommonSubsequence(String a, String b) {
        int alength = a.length() - 1;
        int blength = b.length() - 1;

        if (alength < 0 || blength < 0)
            return "";

        if (a.substring(alength).equals(b.substring(blength))) {
            return longestCommonSubsequence(a.substring(0, alength), b.substring(0, blength))
                    + a.substring(alength);
        } else {
            String first = longestCommonSubsequence(a, b.substring(0, blength));
            String second = longestCommonSubsequence(a.substring(0, alength), b);
            if (first.length() > second.length()) {
                return first;
            } else {
                return second;
            }
        }
    }

答案 1 :(得分:0)

您应该将 a.substring() 更改为 a.charAt()

public static String LCS_r(String a, String b) {
    int m = a.length() - 1;
    int n = b.length() - 1;

    if (m < 0 || n < 0)
        return "";

    if (a.charAt(m)==b.charAt(n)) {
        return LCS_r(a.substring(0, m), b.substring(0, n)) + a.substring(m);
    } 
    else {
        String s1 = LCS_r(a, b.substring(0, n));
        String s2 = LCS_r(a.substring(0, m), b);
        if (s1.length() > s2.length()) {
            return s1;
        } else {
            return s2;
        }
    }