LeetCode - 编辑距离

给定两个单词 word1 和 word2，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：

插入一个字符
删除一个字符
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')

解法1：递归 +备忘录 自顶向下

class Solution:

    memo = {}
    def minDistance(self, word1: str, word2: str) -> int:

        def lwstBT(i , j):

            if (i,j) in self.memo:
                return self.memo[(i,j)]

            if i == -1:
                return j +1
            if j == -1:
                return i+1

            if word1[i] == word2[j]:
                x = lwstBT(i-1, j-1)
            else:
                x =  min(lwstBT(i,j-1) + 1, lwstBT(i-1,j) + 1, lwstBT(i-1,j-1)+1)
            self.memo[(i,j)] = x
            return self.memo[(i,j)]
        self.memo = {}

        return lwstBT(len(word1)-1,len(word2)-1)

解法2：动态规划 自底向上

word1 = "horse"
word2 = "ros"

m  = len(word1)
n = len(word2)

dp = [[0] * (m+1) for i in range(n+1)]

for i in range(1,n+1):

    dp[i][0] = i

for j in range(1,m+1):

    dp[0][j] = j

for i in range(1,m+1):
    for j in range(1,n+1):
        print(dp[j][i])
        if word1[i-1] == word2[j-1]:
            dp[j][i] = dp[j-1][i-1]
        else:
            dp[j][i] = min(dp[j-1][i] + 1,dp[j][i-1]+1,dp[j-1][i-1]+1)

print(dp[-1][-1]