最长公共子序列算法_解决最长公共子序列问题的算法和过程

最长公共子序列算法

最长的公共子序列 (Longest common Subsequence)

  • Let X and Y be two subsequences and by LCS algorithm we can find a maximum length common subsequence of X and Y.

    XY为两个子序列,通过LCS算法,我们可以找到XY最大长度公共子序列

  • If |X|= m and |Y|= n then there is a 2m subsequence of x; we just compare each with Y (n comparisons).

    如果| X | = m| Y | = n,则存在x2m子序列; 我们只将每个与Y比较(n个比较)

  • So the running time of this particular algorithm will be O(n 2m).

    因此,该特定算法的运行时间将为O(n 2m)

  • LCS problem has an optimal substructure which means that the solution of the subproblem is parts of the final solution.

    LCS问题具有最佳子结构,这意味着子问题的解决方案是最终解决方案的一部分。

  • In this case, firstly it’s necessary to find the length of LCS then later we will modify the algorithm to find LCS itself.

    在这种情况下,首先必须找到LCS的长度,然后我们将修改算法以找到LCS本身。

  • For this define Xi, Y¬i to the prefixes of X and Y of length i and j respectively. Then define c[i,j] to be the length of LCS of Xi and Yi.

    为此, 分别X iY¬i定义为长度为ijXY的前缀。 然后将c [i,j]定义为X iY iLCS的长度。

  • Then the length of LCS of X and Y will be c[m, n].

    那么X和YLCS的长度将为c [m,n]

i.e. c[i,j]= c[i-1,j-1]+1 if x[i]=y[j],
      C[i,j]= max(c[i,j-1],c[i-1,j]) otherwise

LCS长度算法 (LCS Length Algorithm)

    1.	m←length[X]
    2.	n←length[y]
    3.	let b[1…m,1…n]and c[0…m,0…n] be new tables
    4.	for i←1 to m
    5.	c[i,j]←0
    6.	do c[0,j]←0
    7.	for j← 0 to n
    8.	do c[0,j]←0
    9.	for i←1 to m
    10.	for j←1 to n
    11.	if xi=yi
    12.	then c[i,j]←c[i-1,j-1]+1
    13.	b[i,j]←”↖”
    14.	else if c[i-1,j]>=c[i,j-1]
    15.	then c[i,j]←c[i-1,j]
    16.	b[i,j]←”↑”
    17.	else c[i,j]←c[i,j-1]
    18.	b[I,j]←”←”
    19.	return c and b

LCS程序 (Procedure of LCS)

  1. X and Y defined as a given X and Y set.

    XY定义为给定的XY集。

  2. First design a, c and b table where Y set values will be present in column side and X set values will be present in row side.

    首先设计acb表,其中Y设置值将出现在列侧,而X设置值将出现在行侧。

  3. Line 1 assigns the length of X into variable m, and in step 2 the length of Y will be assigned to variable n.

    第1行将X的长度分配给变量m ,在步骤2中, Y的长度将分配给变量n

  4. For loop present in line 3 and 4 is used to initialize all the first column value to zero.

    第3和第4行中存在的for循环用于将所有第一列值初始化为零。

  5. Another for loop line 5-6 is used to initialize all the row values to zero.

    另一个for循环行5-6用于将所有行值初始化为零。

  6. In the given algorithm three arrows are used which are vertical, horizontal and semi-vertical. By using these arrows we have to find out the LCS element.

    在给定的算法中,使用了三个箭头,分别是垂直,水平和半垂直。 通过使用这些箭头,我们必须找出LCS元素。

  7. The loop present in line 7-16 is used to insert the numeric values with specific arrows in c and a table.

    本管线7-16环是用来插入在CA表特定箭头数值。

复杂: (Complexity:)

Complexity of longest common subsequence is O(mn). Where, m and n are length of two strings.

最长公共子序列的复杂度为O(mn) 。 其中, mn是两个字符串的长度。

翻译自: https://www.includehelp.com/algorithms/algorithm-and-procedure-to-solve-a-longest-common-subsequence-problem.aspx

最长公共子序列算法

    原文作者:cumubi7453
    原文地址: https://blog.csdn.net/cumubi7453/article/details/107789202
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