583. 两个字符串的删除操作 力扣（LeetCode）官网 - 全球极客挚爱的技术成长平台
提醒
本题和动态规划：115.不同的子序列 相比，其实就是两个字符串都可以删除了，情况虽说复杂一些，但整体思路是不变的
class Solution:
def minDistance(self, word1: str, word2: str) -> int:
dp = [[0] * (len(word2)+1) for _ in range(len(word1)+1)]
for i in range(len(word1)+1):
dp[i][0] = i
for j in range(len(word2)+1):
dp[0][j] = j
for i in range(1, len(word1)+1):
for j in range(1, len(word2)+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] + 2, dp[i-1][j] + 1, dp[i][j-1] + 1)
return dp[-1][-1]72. 编辑距离 力扣（LeetCode）官网 - 全球极客挚爱的技术成长平台
提醒
最终迎来了编辑距离这道题目，之前安排题目都是为了 编辑距离做铺垫
class Solution:
def minDistance(self, word1: str, word2: str) -> int:
dp = [[0] * (len(word2)+1) for _ in range(len(word1)+1)]
for i in range(len(word1)+1):
dp[i][0] = i
for j in range(len(word2)+1):
dp[0][j] = j
for i in range(1, len(word1)+1):
for j in range(1, len(word2)+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-1][j], dp[i][j-1]) + 1
return dp[-1][-1]编辑距离总结篇 代码随想录
一刷没时间，二刷再来
