583. 两个字符串的删除操作

开窍了

class Solution:def minDistance(self, word1: str, word2: str) -> int:n1, n2 = len(word1), len(word2)dp = [[0] * (n2 + 1) for _ in range(n1 + 1)]for i in range(1, n1 + 1):dp[i][0] = ifor j in range(1, n2 + 1):dp[0][j] = jfor i in range(1, n1 + 1):for j in range(1, n2 + 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)return dp[n1][n2]

 72. 编辑距离

因为可以改，dp[i - 1][j -  1]也能直接到dp[i][j]

class Solution:def minDistance(self, word1: str, word2: str) -> int:n1, n2 = len(word1), len(word2)dp = [[0] * (n2 + 1) for _ in range(n1 + 1)]for i in range(1, n1 + 1):dp[i][0] = ifor j in range(1, n2 + 1):dp[0][j] = jfor i in range(1, n1 + 1):for j in range(1, n2 + 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[n1][n2]

今日总结：

继昨天，理解了“删除”的含义了。

其实和可以从后向前思考，要得到dp[i][j]，dp[i-1] dp[j - 1] 都要做些什么。