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! |
