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[feat] add ruby code - chapter dynamic programming (#1378)
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115
codes/ruby/chapter_dynamic_programming/edit_distance.rb
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115
codes/ruby/chapter_dynamic_programming/edit_distance.rb
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=begin
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File: edit_distance.rb
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Created Time: 2024-05-29
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Author: Xuan Khoa Tu Nguyen (ngxktuzkai2000@gmail.com)
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=end
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### 编辑距离:暴力搜索 ###
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def edit_distance_dfs(s, t, i, j)
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# 若 s 和 t 都为空,则返回 0
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return 0 if i == 0 && j == 0
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# 若 s 为空,则返回 t 长度
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return j if i == 0
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# 若 t 为空,则返回 s 长度
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return i if j == 0
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# 若两字符相等,则直接跳过此两字符
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return edit_distance_dfs(s, t, i - 1, j - 1) if s[i - 1] == t[j - 1]
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# 最少编辑步数 = 插入、删除、替换这三种操作的最少编辑步数 + 1
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insert = edit_distance_dfs(s, t, i, j - 1)
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delete = edit_distance_dfs(s, t, i - 1, j)
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replace = edit_distance_dfs(s, t, i - 1, j - 1)
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# 返回最少编辑步数
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[insert, delete, replace].min + 1
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end
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def edit_distance_dfs_mem(s, t, mem, i, j)
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# 若 s 和 t 都为空,则返回 0
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return 0 if i == 0 && j == 0
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# 若 s 为空,则返回 t 长度
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return j if i == 0
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# 若 t 为空,则返回 s 长度
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return i if j == 0
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# 若已有记录,则直接返回之
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return mem[i][j] if mem[i][j] != -1
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# 若两字符相等,则直接跳过此两字符
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return edit_distance_dfs_mem(s, t, mem, i - 1, j - 1) if s[i - 1] == t[j - 1]
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# 最少编辑步数 = 插入、删除、替换这三种操作的最少编辑步数 + 1
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insert = edit_distance_dfs_mem(s, t, mem, i, j - 1)
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delete = edit_distance_dfs_mem(s, t, mem, i - 1, j)
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replace = edit_distance_dfs_mem(s, t, mem, i - 1, j - 1)
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# 记录并返回最少编辑步数
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mem[i][j] = [insert, delete, replace].min + 1
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end
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### 编辑距离:动态规划 ###
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def edit_distance_dp(s, t)
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n, m = s.length, t.length
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dp = Array.new(n + 1) { Array.new(m + 1, 0) }
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# 状态转移:首行首列
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(1...(n + 1)).each { |i| dp[i][0] = i }
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(1...(m + 1)).each { |j| dp[0][j] = j }
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# 状态转移:其余行和列
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for i in 1...(n + 1)
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for j in 1...(m +1)
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if s[i - 1] == t[j - 1]
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# 若两字符相等,则直接跳过此两字符
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dp[i][j] = dp[i - 1][j - 1]
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else
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# 最少编辑步数 = 插入、删除、替换这三种操作的最少编辑步数 + 1
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dp[i][j] = [dp[i][j - 1], dp[i - 1][j], dp[i - 1][j - 1]].min + 1
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end
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end
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end
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dp[n][m]
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end
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### 编辑距离:空间优化后的动态规划 ###
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def edit_distance_dp_comp(s, t)
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n, m = s.length, t.length
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dp = Array.new(m + 1, 0)
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# 状态转移:首行
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(1...(m + 1)).each { |j| dp[j] = j }
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# 状态转移:其余行
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for i in 1...(n + 1)
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# 状态转移:首列
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leftup = dp.first # 暂存 dp[i-1, j-1]
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dp[0] += 1
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# 状态转移:其余列
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for j in 1...(m + 1)
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temp = dp[j]
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if s[i - 1] == t[j - 1]
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# 若两字符相等,则直接跳过此两字符
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dp[j] = leftup
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else
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# 最少编辑步数 = 插入、删除、替换这三种操作的最少编辑步数 + 1
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dp[j] = [dp[j - 1], dp[j], leftup].min + 1
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end
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leftup = temp # 更新为下一轮的 dp[i-1, j-1]
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end
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end
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dp[m]
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end
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### Driver Code ###
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if __FILE__ == $0
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s = 'bag'
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t = 'pack'
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n, m = s.length, t.length
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# 暴力搜索
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res = edit_distance_dfs(s, t, n, m)
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puts "将 #{s} 更改为 #{t} 最少需要编辑 #{res} 步"
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# 记忆化搜索
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mem = Array.new(n + 1) { Array.new(m + 1, -1) }
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res = edit_distance_dfs_mem(s, t, mem, n, m)
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puts "将 #{s} 更改为 #{t} 最少需要编辑 #{res} 步"
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# 动态规划
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res = edit_distance_dp(s, t)
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puts "将 #{s} 更改为 #{t} 最少需要编辑 #{res} 步"
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# 空间优化后的动态规划
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res = edit_distance_dp_comp(s, t)
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puts "将 #{s} 更改为 #{t} 最少需要编辑 #{res} 步"
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end
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