[算法练习]最长公共子序列

题目说明：

找两个字符串的最长公共子序列，这个子序列不要求在原字符串中是连续的，但要求顺序是一致的，比如"abcd"和"aebfc"的最长公共子串是"abc"。

程序代码：

#include <gtest/gtest.h>
#include <vector>
using namespace std;

enum DirectionType
{
    TypeUp,
    TypeLeftUp,
    TypeLeft
};

int TMaxVal(int A, int B, int C, DirectionType& index)
{
    int nVal = A;
    index = TypeUp;

    if (B > nVal)
    {
        nVal = B;
        index = TypeLeftUp;
    }

    if (C > nVal)
    {
        nVal = C;
        index = TypeLeft;
    }

    return nVal;
}

// 按照上、左上、左的比较顺序
string GetLCS(const string& strA, const string& strB)
{
    int nLengthA = strA.length();
    int nLengthB = strB.length();

    if (!nLengthA || !nLengthB)
    {
        return "";
    }

    struct RecordInfo
    {
        char Direction;
        char Count;
    };

    RecordInfo* arrRecordInfo = new RecordInfo[(nLengthA + 1) * (nLengthB+1)];
    memset(arrRecordInfo, 0, (nLengthA + 1) * (nLengthB + 1) * sizeof(RecordInfo));

    for (int i = 1; i <= nLengthA; ++i)
    {
        for (int j = 1; j <= nLengthB; ++j)
        {
            int nNewVal = 0;
            DirectionType nDirect;
            if (strA[i-1] == strB[j-1])
            {
                nNewVal = TMaxVal(
                    arrRecordInfo[(i-1)*(nLengthB+1) + j].Count,
                    arrRecordInfo[(i-1)*(nLengthB+1) + j-1].Count + 1,
                    arrRecordInfo[i*(nLengthB+1) + j-1].Count,
                    nDirect);
            }
            else
            {
                nNewVal = TMaxVal(
                    arrRecordInfo[(i-1)*(nLengthB+1) + j].Count,
                    arrRecordInfo[(i-1)*(nLengthB+1) + j-1].Count,
                    arrRecordInfo[i*(nLengthB+1) + j-1].Count,
                    nDirect);
            }

            arrRecordInfo[i*(nLengthB+1) + j].Count = nNewVal;
            arrRecordInfo[i*(nLengthB+1) + j].Direction = nDirect;
        }
    }

    string strValue;
    int i = nLengthA;
    int j = nLengthB;
    while ((i > 0) && (j > 0))
    {
        int nDirect = arrRecordInfo[i*(nLengthB+1) + j].Direction;
        if (TypeLeftUp == nDirect)
        {
            if (strA[i-1] == strB[j-1])
            {
                strValue += strA[i-1];
            }

            --i;
            --j;
        }
        else if(TypeUp == nDirect)
        {
            --i;
        }
        else if(TypeLeft == nDirect)
        {
            --j;
        }
    }

    reverse(strValue.begin(),strValue.end());

    return strValue;
}

TEST(Algo, tLongestCommonSequence)
{
    //
    // "" ""
    // abcdefg   bdacafg
    // abcabcagc abcdcagb

    vector<string> result;
    ASSERT_EQ(GetLCS("",""),"");
    ASSERT_EQ(GetLCS("abcdefg","bdacafg"), "acfg");
}