最长的共同子序列

时间:2013-10-19 22:33:42

标签: java recursion infinite lcs

我正在尝试编写一个递归算法,找到两个列表中最长的公共子序列,如http://en.wikipedia.org/wiki/Longest_common_subsequence_problem#LCS_function_defined

中所述

似乎递归永远不会结束,我无法弄清楚我正在做什么呢

public static List<ActionType> getLongestSequence(List<ActionType> list1, List<ActionType> list2) {
    return getLongestSequence(list1, list2, list1.size(), list2.size());
}

public static List<ActionType> getLongestSequence(List<ActionType> list1, List<ActionType> list2, int list1index, int list2index) {

    if (list1index == 0 || list2index == 0) {
        return new ArrayList<ActionType>();
    }

    if (list1.get(list1index-1).equals(list2.get(list2index-1))) {
        List<ActionType> retVal = getLongestSequence(list1, list2, list1index-1, list2index-1);
        retVal.add(list1.get(list1index-1));
        return retVal;
    } else {
        List<ActionType> ret1 = getLongestSequence(list1, list2, list1index, list2index-1);
        List<ActionType> ret2 = getLongestSequence(list1, list2, list1index-1, list2index);

        if (ret1.size() > ret2.size()) {
            return ret1;
        } else {
            return ret2;
        }
    }
}

任何帮助解决这个问题的人都表示赞赏。谢谢。

1 个答案:

答案 0 :(得分:1)

这个问题很复杂。实现记忆将运行时间从一天减少到几秒钟。

以下是更新的算法:

public static List<ActionType> getLongestSequence(List<ActionType> list1, List<ActionType> list2) {
    lcsMemorize = new HashMap<Integer, List<ActionType>>();
    return getLongestSequence(list1, list2, list1.size(), list2.size());
}

public static List<ActionType> getLongestSequence(List<ActionType> list1, List<ActionType> list2, int list1index, int list2index) {

    List<ActionType> retVal = lcsMemorize.get(list1index + list2index * 1000000);

    if (retVal != null) {
        return retVal;
    } else if (list1index == 0 || list2index == 0) {
        retVal = new ArrayList<ActionType>();
    } else if (list1.get(list1index-1).equals(list2.get(list2index-1))) {
        List<ActionType> returned = getLongestSequence(list1, list2, list1index-1, list2index-1);

        retVal = new ArrayList<ActionType>(returned);
        retVal.add(list1.get(list1index-1));
    } else {
        List<ActionType> ret1 = getLongestSequence(list1, list2, list1index, list2index-1);
        List<ActionType> ret2 = getLongestSequence(list1, list2, list1index-1, list2index);

        if (ret1.size() > ret2.size()) {
            retVal = ret1;
        } else {
            retVal = ret2;
        }
    }

    lcsMemorize.put(list1index + list2index * 1000000, retVal);

    return retVal;
}

注意:

在我的运行中,原始列表长度为1100 - 1300个元素,ActionType是枚举。这种方法使用了大量内存。我不得不将JVM堆大小增加到4GB。