最大匹配字符串LCS,The Longest Common Substring

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public enum BackTracking
{
    UP,
    LEFT,
    NEITHER,
    UP_AND_LEFT
}
 
public abstract class LCSBaseMatch
{
    /// <summary>
    /// 设置连续字符的匹配值
    /// </summary>
    /// <param name="length"></param>
    /// <returns></returns>
    protected virtual int ConsecutiveMeasure(int length)
    {
        return length * length;
    }
 
    /// <summary>
    /// 获取两个string字符串的匹配度
    /// </summary>
    /// <param name="list1"></param>
    /// <param name="list2"></param>
    /// <returns></returns>
    public virtual int GetMatchScoreOfLCS(char[] list1, char[] list2)
    {
        int[,] lcs;//最大匹配度
        BackTracking[,] backTracer;//需要执行的操作
 
        int score = GetMatchScoreOfLCS(list1, list2, out lcs, out backTracer);//最终匹配度
        return score;
    }
 
    /// <summary>
    /// 计算匹配度
    /// </summary>
    /// <param name="list1"></param>
    /// <param name="list2"></param>
    /// <param name="lcs"></param>
    /// <param name="backTracer"></param>
    /// <returns></returns>
    protected int GetMatchScoreOfLCS(char[] list1, char[] list2, out int[,] lcs, out BackTracking[,] backTracer)
    {
        int m = list1.Length;
        int n = list2.Length;
 
        lcs = new int[m, n];//最大匹配度
        backTracer = new BackTracking[m, n];//需要执行的操作
        int[,] w = new int[m, n];//连续匹配的长度
        int i, j;
 
        #region 初始化lcs、backTracer
        for (i = 0; i < m; ++i)
        {
            lcs[i, 0] = 0;
            backTracer[i, 0] = BackTracking.UP;
        }
        for (j = 0; j < n; ++j)
        {
            lcs[0, j] = 0;
            backTracer[0, j] = BackTracking.LEFT;
        }
        #endregion
 
        #region 给lcs、backTracer、w赋值
        for (i = 0; i < m; i++)
        {
            for (j = 0; j < n; j++)
            {
                if (list1[i] == list2[j])
                {
                    int k = 0;
                    int prev = 0;
                    if (i > 0 && j > 0)
                    {
                        k = w[i - 1, j - 1];
                        prev = lcs[i - 1, j - 1];
                    }
                    //eviation unit between k+1 and k instead of 1 in basic LCS
                    lcs[i, j] = prev + ConsecutiveMeasure(k + 1) - ConsecutiveMeasure(k);
                    backTracer[i, j] = BackTracking.UP_AND_LEFT;
                    w[i, j] = k + 1;
                }
                if (i > 0 && (lcs[i - 1, j] > lcs[i, j]))
                {
                    lcs[i, j] = lcs[i - 1, j];
                    backTracer[i, j] = BackTracking.UP;
                    w[i, j] = 0;
                }
                if (j > 0 && (lcs[i, j - 1] > lcs[i, j]))
                {
                    lcs[i, j] = lcs[i, j - 1];
                    backTracer[i, j] = BackTracking.LEFT;
                    w[i, j] = 0;
                }
            }
        }
        #endregion
 
        return lcs[m - 1, n - 1];//最终匹配度
    }
 
}
 
 
public class LCSMatchForString : LCSBaseMatch
{
    /// <summary>
    /// get The Longest Common Substring
    /// </summary>
    /// <param name="list1"></param>
    /// <param name="list2"></param>
    /// <returns></returns>
    public string LCS(string s1, string s2)
    {
        char[] list1 = s1.ToArray();
        char[] list2 = s2.ToArray();
        int m = list1.Length;
        int n = list2.Length;
 
        int[,] lcs ;//最大匹配度
        BackTracking[,] backTracer ;//需要执行的操作
 
        int score = GetMatchScoreOfLCS(list1,list2,out lcs,out backTracer);//最终匹配度
 
        #region 获取最大匹配的字符串
        int i = m - 1;
        int j = n - 1;
        string subseq = "";
        //trace the backtracking matrix.
        while (i >=0 && j >=0)
        {
            if (backTracer[i, j] == BackTracking.NEITHER) break;
            if (backTracer[i, j] == BackTracking.UP_AND_LEFT)
            {
                subseq = list1[i]+ subseq;
                i--;
                j--;
            }
            else if (backTracer[i, j] == BackTracking.UP)
            {
                i--;
            }
            else if (backTracer[i, j] == BackTracking.LEFT)
            {
                j--;
            }
        }
        #endregion
 
        return subseq;
    }   
}

  

posted @   PanPan003  阅读(573)  评论(0编辑  收藏  举报
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