583. 两个字符串的删除操作
# 给定两个单词word1和word2，返回使得word1和word2相同所需的最小步数。 # 每步可以删除任意一个字符串中的一个字符。 # # 示例 1： # 输入: word1 = "sea", word2 = "eat" # 输出: 2 # 解释: 第一步将 "sea" 变为 "ea" ，第二步将 "eat "变为 "ea" # # 示例 2: # 输入：word1 = "leetcode", word2 = "etco" # 输出：4 # # 提示： # 1 <= word1.length, word2.length <= 500 # word1和word2只包含小写英文字母
class Solution:
def minDistance(self, word1: str, word2: str) -> int:
# 听不懂啊
# 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
# 给你两个单词 word1 和 word2， 请返回将 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') # # 提示： # 0 <= word1.length, word2.length <= 500 # word1 和 word2 由小写英文字母组成
class Solution:
def minDistance(self, word1: str, word2: str) -> int:
# 啊啊啊啊啊啊，痛苦啊
# 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
