文件名称:LCSProblem
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LCS算法:
通常两个字符串的最大公共子串的问题是通过下面的算法来完成的： 把字符串1（长度m）横排，串2（长度n）竖排，得到一个m×n的矩阵c，矩阵的每个元素的值如下，如果m[i]=n[j]，则c[j][i]=1,否则，c[j][i]=0。然后找出矩阵中连续是1的对角线最长的一个，则对角线的长度就是公共子串的长度.-LCS algorithm: often the two most common sub-string string problem is through the following algorithm to accomplish: the string 1 (length m) horizontal, series 2 (length n) vertical, get an m × n of matrix c, the value of each element matrix is as follows, if m [i] = n [j], then c [j] [i] = 1, otherwise, c [j] [i] = 0. Then find the matrix is a continuous one of the longest diagonal, then the diagonal length of the substring is the length of the public.
通常两个字符串的最大公共子串的问题是通过下面的算法来完成的： 把字符串1（长度m）横排，串2（长度n）竖排，得到一个m×n的矩阵c，矩阵的每个元素的值如下，如果m[i]=n[j]，则c[j][i]=1,否则，c[j][i]=0。然后找出矩阵中连续是1的对角线最长的一个，则对角线的长度就是公共子串的长度.-LCS algorithm: often the two most common sub-string string problem is through the following algorithm to accomplish: the string 1 (length m) horizontal, series 2 (length n) vertical, get an m × n of matrix c, the value of each element matrix is as follows, if m [i] = n [j], then c [j] [i] = 1, otherwise, c [j] [i] = 0. Then find the matrix is a continuous one of the longest diagonal, then the diagonal length of the substring is the length of the public.
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LCSProblem.txt