Ruby實現的最長公用子序列演算法

Ruby實現的最長公用子序列演算法

 　　這篇文章主要介紹了Ruby實現的最長公用子序列演算法,本文直接給出實現代碼,需要的朋友可以參考下

　　最長公用子序列，LCS，動態規劃實現。

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#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", "a", "a", "b", "d"] and ["a", "c", "a", "c", "a", "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 %s\n", 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, b, i-1, j-1) printf("%c ", x[i - 1]) elsif(b[i][j] == 1) Print_LCS(x, b, i-1, j) else Print_LCS(x, b, i, j-1) end end   LCS_length(x, y, c ,b) print "The Longest-Commom-Subsequence is: " Print_LCS(x, b, x.size, y.size)