mirror of
https://github.com/SunnyQjm/algorithm-review.git
synced 2026-06-03 08:16:43 +08:00
86 lines
2.9 KiB
Python
86 lines
2.9 KiB
Python
#!/usr/bin/env python
|
||
# coding=utf-8
|
||
|
||
#######################################################################################
|
||
# Leetcode 72 编辑距离
|
||
#
|
||
# 给你两个单词 word1 和 word2，请你计算出将 word1 转换成 word2 所使用的最少操作数 。
|
||
# 你可以对一个单词进行如下三种操作：
|
||
# 1. 插入一个字符
|
||
# 2. 删除一个字符
|
||
# 3. 替换一个字符
|
||
#
|
||
# 示例 1：
|
||
# 输入：word1 = "horse", word2 = "ros"
|
||
# 输出：3
|
||
# 解释：
|
||
# horse -> rorse (将 'h' 替换为 'r')
|
||
# rorse -> rose (删除 'r')
|
||
# rose -> ros (删除 'e')
|
||
#
|
||
# 示例 2：
|
||
# 输入：word1 = "intention", word2 = "execution"
|
||
# 输出：5
|
||
# 解释：
|
||
# intention -> inention (删除 't')
|
||
# inention -> enention (将 'i' 替换为 'e')
|
||
# enention -> exention (将 'n' 替换为 'x')
|
||
# exention -> exection (将 'n' 替换为 'c')
|
||
# exection -> execution (插入 'u')
|
||
#######################################################################################
|
||
|
||
|
||
class Solution:
|
||
def minDistance(self, word1, word2):
|
||
"""
|
||
:type word1: str
|
||
:type word2: str
|
||
:rtype int
|
||
|
||
(knowledge)
|
||
|
||
思路：
|
||
其实总共有四种操作，插入、删除、替换和什么都不操作
|
||
1. 使用动态规划；
|
||
2. 定义状态：dp[i][j] => word1前i个字符构成的子串与word2前j个字符构成的子串的编辑距离；
|
||
3. base case => dp[i][0] = i, dp[0][j] = j => 表示其中一个子串为空的情形
|
||
4. 状态转移方程：
|
||
f(i, j) = j i == 0
|
||
i j == 0
|
||
f(i - 1, j - 1) i > 0 && j > 0 && word1[i] == word2[j]
|
||
min { i > 0 && j > 0 && word1[i] != word2[j]
|
||
f(i - 1, j) + 1,
|
||
f(i, j - 1) + 1,
|
||
f(i - 1, j - 1) + 1
|
||
}
|
||
|
||
tip: 可以参考 => https://labuladong.gitbook.io/algo/dong-tai-gui-hua-xi-lie/bian-ji-ju-li
|
||
"""
|
||
m, n = len(word1), len(word2)
|
||
dp = [[0] * (n + 1) for i in range(m + 1)]
|
||
|
||
for i in range(1, m + 1):
|
||
dp[i][0] = i
|
||
|
||
for j in range(1, n + 1):
|
||
dp[0][j] = j
|
||
|
||
for i in range(1, m + 1):
|
||
for j in range(1, n + 1):
|
||
if word1[i - 1] == word2[j - 1]:
|
||
dp[i][j] = dp[i - 1][j - 1]
|
||
else:
|
||
dp[i][j] = min(
|
||
dp[i - 1][j] + 1,
|
||
dp[i][j - 1] + 1,
|
||
dp[i - 1][j - 1] + 1
|
||
)
|
||
|
||
return dp[-1][-1]
|
||
|
||
|
||
if __name__ == '__main__':
|
||
solution = Solution()
|
||
print(solution.minDistance("horse", "ros"), "= 3")
|
||
print(solution.minDistance("intention", "execution"), "= 5")
|