Ruby实现的最长公共子序列算法
发布时间:2020-12-16 23:36:12 所属栏目:百科 来源:网络整理
导读:最长公共子序列,LCS,动态规划实现。 #encoding: utf-8#author: xu jin,4100213#date: Nov 01,2012#Longest-Commom-Subsequence#to find a longest commom subsequence of two given character arrays by using LCS algorithm#example output:#The random c
最长公共子序列,LCS,动态规划实现。 #encoding: utf-8 #author: xu jin,4100213 #date: Nov 01,2012 #Longest-Commom-Subsequence #to find a longest commom subsequence of two given character arrays by using LCS algorithm #example output: #The random character arrays are: ["b","a","c","b","d"] and ["a","b"] #The Longest-Commom-Subsequence is: a c a a b chars = ("a".."e").to_a x,y = [],[] 1.upto(rand(5) + 5) { |i| x << chars[rand(chars.size-1)] } 1.upto(rand(5) + 5) { |i| y << chars[rand(chars.size-1)] } printf("The random character arrays are: %s and %sn",x,y) c = Array.new(x.size + 1){Array.new(y.size + 1)} b = Array.new(x.size + 1){Array.new(y.size + 1)} def LCS_length(x,y,c,b) m,n = x.size,y.size (0..m).each{|i| c[i][0] = 0} (0..n).each{|j| c[0][j] = 0} for i in (1..m) do for j in(1..n) do if(x[i - 1] == y [j - 1]) c[i][j] = c[i - 1][j - 1] + 1; b[i][j] = 0 else if(c[i - 1][j] >= c[i][j - 1]) c[i][j] = c[i - 1][j] b[i][j] = 1 else c[i][j] = c[i][j - 1] b[i][j] = 2 end end end end end def Print_LCS(x,b,i,j) return if(i == 0 || j == 0) if(b[i][j] == 0) Print_LCS(x,i-1,j-1) printf("%c ",x[i - 1]) elsif(b[i][j] == 1) Print_LCS(x,j) else Print_LCS(x,j-1) end end LCS_length(x,b) print "The Longest-Commom-Subsequence is: " Print_LCS(x,x.size,y.size) (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |