最長公共子序列(Longest common subsequence)

問題描述：

    給定兩個序列 X=<x1, x2, ..., xm>, Y<y1, y2, ..., yn>，求X和Y長度最長的公共子序列。(子序列中的字符不要求連續)

 

   這道題可以用動態規划解決。定義c[i, j]表示Xi和Yj的LCS的長度，可得如下公式：

偽代碼如下：

C++實現：

int longestCommonSubsequence(string x, string y)
{
    int m = x.length();
    int n = y.length();
    vector< vector<int> > c(m + 1, vector<int>(n + 1));

    for (int i = 0; i <= m; ++i)
        c[i][0] = 0;
    for (int j = 1; j <= n; ++j)
        c[0][j] = 0;
    for (int i = 1; i <= m; ++i)
    {
        for (int j = 1; j <= n; ++j)
        {
            if (x[i-1] == y[j-1])
                c[i][j] = c[i-1][j-1] + 1;
            else if (c[i-1][j] >= c[i][j-1])
                c[i][j] = c[i-1][j];
            else
                c[i][j] = c[i][j-1];
        }
    }
    return c[m][n];
}

后記：

我本來以為我已經掌握了LCS，其實不過是記住了LCS的狀態轉移方程。15號參加了創新工場2016校園招聘筆試，題目要求打印出LCS，我就懵逼了。其實《算法導論》里講的清清楚楚啊。

 

貼一下我的C++實現：

vector< vector<int> > b;    //輔助數組
void LCS(string x, string y)
{
    int m = x.length();
    int n = y.length();

    vector< vector<int> > c(m + 1, vector<int>(n + 1));
    for (int i = 0; i <= m; ++i)
        c[i][0] = 0;
    for (int j = 1; j <= n; ++j)
        c[0][j] = 0;

    b.resize(m+1);
    for (int i = 1; i <= m; i++)
    {
        b[i].resize(n+1);
    }
    for (int i = 1; i <= m; i++)
        for (int j = 1; j <= n; j++)
        {
            b[i][j] = 0;
        }

    for (int i = 1; i <= m; ++i)
    {
        for (int j = 1; j <= n; ++j)
        {
            if (x[i-1] == y[j-1])
            {
                c[i][j] = c[i-1][j-1] + 1;
                b[i][j] = 1;    //
            }
            else if (c[i-1][j] >= c[i][j-1])
            {
                c[i][j] = c[i-1][j];
                b[i][j] = 2;   //
            }
            else
            {
                c[i][j] = c[i][j-1];
                b[i][j] = 3;   //
            }
        }
    }

}

void printLCS(vector< vector<int> > &b, string x, int i, int j)
{
    if (i == 0 || j == 0)
        return ;
    if (b[i][j] == 1)
    {
        printLCS(b, x, i-1, j-1);
        printf("%c", x[i-1]);
    }
    else if (b[i][j] == 2)
        printLCS(b, x, i-1, j);
    else
        printLCS(b, x, i, j-1);
    
}

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