Leetcode刷题笔记1-动态规划part12
目录
Leetcode刷题笔记1 动态规划part12
leetcode 115 不同的子序列
在子序列问题中，遇到不匹配的情况时，哪个子序列长，哪个子序列让步，如果不知道哪个长就直接max(dp[i - 1][j], dp[i][j - 1])
class Solution:
def numDistinct(self, s: str, t: str) -> int:
dp = [[0] * (len(t) + 1) for _ in range(len(s) + 1)]
for i in range(len(s) + 1):
dp[i][0] = 1
for i in range(1, len(s) + 1):
for j in range(1, len(t) + 1):
if s[i - 1] == t[j - 1]:
dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j]
else:
dp[i][j] = dp[i - 1][j]
return dp[-1][-1]leetcode 583 两个字符串的删除操作
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(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] + 1
else:
dp[i][j] = max(dp[i][j - 1], dp[i - 1][j])
return len(word1) + len(word2) - 2 * dp[-1][-1]leetcode 72 编辑距离
在考虑不等情况时，删除和增加等同，替换是在dp[i - 1][j - 1]的基础上加一，这点很关键
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]