""" File: edit_distancde.py Created Time: 2023-07-04 Author: krahets (krahets@163.com) """ def edit_distance_dfs(s: str, t: str, i: int, j: int) -> int: """编辑距离:暴力搜索""" # 若 s 和 t 都为空,则返回 0 if i == 0 and j == 0: return 0 # 若 s 为空,则返回 t 长度 if i == 0: return j # 若 t 为空,则返回 s 长度 if j == 0: return i # 若两字符相等,则直接跳过此两字符 if s[i - 1] == t[j - 1]: return edit_distance_dfs(s, t, i - 1, j - 1) # 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 insert = edit_distance_dfs(s, t, i, j - 1) delete = edit_distance_dfs(s, t, i - 1, j) replace = edit_distance_dfs(s, t, i - 1, j - 1) # 返回最少编辑步数 return min(insert, delete, replace) + 1 def edit_distance_dfs_mem(s: str, t: str, mem: list[list[int]], i: int, j: int) -> int: """编辑距离:记忆化搜索""" # 若 s 和 t 都为空,则返回 0 if i == 0 and j == 0: return 0 # 若 s 为空,则返回 t 长度 if i == 0: return j # 若 t 为空,则返回 s 长度 if j == 0: return i # 若已有记录,则直接返回之 if mem[i][j] != -1: return mem[i][j] # 若两字符相等,则直接跳过此两字符 if s[i - 1] == t[j - 1]: return edit_distance_dfs_mem(s, t, mem, i - 1, j - 1) # 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 insert = edit_distance_dfs_mem(s, t, mem, i, j - 1) delete = edit_distance_dfs_mem(s, t, mem, i - 1, j) replace = edit_distance_dfs_mem(s, t, mem, i - 1, j - 1) # 记录并返回最少编辑步数 mem[i][j] = min(insert, delete, replace) + 1 return mem[i][j] def edit_distance_dp(s: str, t: str) -> int: """编辑距离:动态规划""" n, m = len(s), len(t) dp = [[0] * (m + 1) for _ in range(n + 1)] # 状态转移:首行首列 for i in range(1, n + 1): dp[i][0] = i for j in range(1, m + 1): dp[0][j] = j # 状态转移:其余行和列 for i in range(1, n + 1): for j in range(1, m + 1): if s[i - 1] == t[j - 1]: # 若两字符相等,则直接跳过此两字符 dp[i][j] = dp[i - 1][j - 1] else: # 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 dp[i][j] = min(dp[i][j - 1], dp[i - 1][j], dp[i - 1][j - 1]) + 1 return dp[n][m] def edit_distance_dp_comp(s: str, t: str) -> int: """编辑距离:空间优化后的动态规划""" n, m = len(s), len(t) dp = [0] * (m + 1) # 状态转移:首行 for j in range(1, m + 1): dp[j] = j # 状态转移:其余行 for i in range(1, n + 1): # 状态转移:首列 leftup = dp[0] # 暂存 dp[i-1, j-1] dp[0] += 1 # 状态转移:其余列 for j in range(1, m + 1): temp = dp[j] if s[i - 1] == t[j - 1]: # 若两字符相等,则直接跳过此两字符 dp[j] = leftup else: # 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 dp[j] = min(dp[j - 1], dp[j], leftup) + 1 leftup = temp # 更新为下一轮的 dp[i-1, j-1] return dp[m] """Driver Code""" if __name__ == "__main__": s = "bag" t = "pack" n, m = len(s), len(t) # 暴力搜索 res = edit_distance_dfs(s, t, n, m) print(f"将 {s} 更改为 {t} 最少需要编辑 {res} 步") # 记忆化搜索 mem = [[-1] * (m + 1) for _ in range(n + 1)] res = edit_distance_dfs_mem(s, t, mem, n, m) print(f"将 {s} 更改为 {t} 最少需要编辑 {res} 步") # 动态规划 res = edit_distance_dp(s, t) print(f"将 {s} 更改为 {t} 最少需要编辑 {res} 步") # 空间优化后的动态规划 res = edit_distance_dp_comp(s, t) print(f"将 {s} 更改为 {t} 最少需要编辑 {res} 步")