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124 lines
4.1 KiB
124 lines
4.1 KiB
"""
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File: edit_distancde.py
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Created Time: 2023-07-04
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Author: krahets (krahets@163.com)
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"""
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def edit_distance_dfs(s: str, t: str, i: int, j: int) -> int:
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"""編輯距離：暴力搜尋"""
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# 若 s 和 t 都為空，則返回 0
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if i == 0 and j == 0:
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return 0
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# 若 s 為空，則返回 t 長度
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if i == 0:
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return j
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# 若 t 為空，則返回 s 長度
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if j == 0:
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return i
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# 若兩字元相等，則直接跳過此兩字元
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if s[i - 1] == t[j - 1]:
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return edit_distance_dfs(s, t, i - 1, j - 1)
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# 最少編輯步數 = 插入、刪除、替換這三種操作的最少編輯步數 + 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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# 返回最少編輯步數
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return min(insert, delete, replace) + 1
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def edit_distance_dfs_mem(s: str, t: str, mem: list[list[int]], i: int, j: int) -> int:
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"""編輯距離：記憶化搜尋"""
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# 若 s 和 t 都為空，則返回 0
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if i == 0 and j == 0:
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return 0
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# 若 s 為空，則返回 t 長度
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if i == 0:
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return j
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# 若 t 為空，則返回 s 長度
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if j == 0:
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return i
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# 若已有記錄，則直接返回之
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if mem[i][j] != -1:
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return mem[i][j]
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# 若兩字元相等，則直接跳過此兩字元
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if s[i - 1] == t[j - 1]:
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return edit_distance_dfs_mem(s, t, mem, i - 1, j - 1)
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# 最少編輯步數 = 插入、刪除、替換這三種操作的最少編輯步數 + 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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# 記錄並返回最少編輯步數
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mem[i][j] = min(insert, delete, replace) + 1
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return mem[i][j]
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def edit_distance_dp(s: str, t: str) -> int:
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"""編輯距離：動態規劃"""
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n, m = len(s), len(t)
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dp = [[0] * (m + 1) for _ in range(n + 1)]
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# 狀態轉移：首行首列
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for i in range(1, n + 1):
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dp[i][0] = i
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for j in range(1, m + 1):
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dp[0][j] = j
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# 狀態轉移：其餘行和列
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for i in range(1, n + 1):
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for j in range(1, m + 1):
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if s[i - 1] == t[j - 1]:
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# 若兩字元相等，則直接跳過此兩字元
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dp[i][j] = dp[i - 1][j - 1]
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else:
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# 最少編輯步數 = 插入、刪除、替換這三種操作的最少編輯步數 + 1
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dp[i][j] = min(dp[i][j - 1], dp[i - 1][j], dp[i - 1][j - 1]) + 1
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return dp[n][m]
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def edit_distance_dp_comp(s: str, t: str) -> int:
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"""編輯距離：空間最佳化後的動態規劃"""
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n, m = len(s), len(t)
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dp = [0] * (m + 1)
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# 狀態轉移：首行
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for j in range(1, m + 1):
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dp[j] = j
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# 狀態轉移：其餘行
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for i in range(1, n + 1):
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# 狀態轉移：首列
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leftup = dp[0] # 暫存 dp[i-1, j-1]
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dp[0] += 1
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# 狀態轉移：其餘列
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for j in range(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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# 若兩字元相等，則直接跳過此兩字元
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dp[j] = leftup
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else:
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# 最少編輯步數 = 插入、刪除、替換這三種操作的最少編輯步數 + 1
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dp[j] = min(dp[j - 1], dp[j], leftup) + 1
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leftup = temp # 更新為下一輪的 dp[i-1, j-1]
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return dp[m]
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"""Driver Code"""
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if __name__ == "__main__":
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s = "bag"
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t = "pack"
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n, m = len(s), len(t)
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# 暴力搜尋
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res = edit_distance_dfs(s, t, n, m)
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print(f"將 {s} 更改為 {t} 最少需要編輯 {res} 步")
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# 記憶化搜尋
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mem = [[-1] * (m + 1) for _ in range(n + 1)]
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res = edit_distance_dfs_mem(s, t, mem, n, m)
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print(f"將 {s} 更改為 {t} 最少需要編輯 {res} 步")
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# 動態規劃
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res = edit_distance_dp(s, t)
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print(f"將 {s} 更改為 {t} 最少需要編輯 {res} 步")
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# 空間最佳化後的動態規劃
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res = edit_distance_dp_comp(s, t)
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print(f"將 {s} 更改為 {t} 最少需要編輯 {res} 步")
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