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72. Edit Distance
f(i, j) := minimum cost (or steps) required to convert first i characters of word1 to first j characters of word2
class Solution(object):
def minDistance(self, word1, word2):
"""
:type word1: str
:type word2: str
:rtype: int
"""
n1, n2 = len(word1), len(word2)
dp = [[0] * (n2+1) for _ in xrange(n1+1)]
for i in xrange(n1+1):
dp[i][0] = i
for j in xrange(n2+1):
dp[0][j] = j
for i in xrange(1, n1+1):
for j in xrange(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], dp[i][j-1], dp[i-1][j-1]) + 1
return dp[-1][-1]
97. Interleaving String
丢人啊，这题不难第一次解又没有想到AC的解法，又是超时。
DP：
class Solution(object):
def isInterleave(self, s1, s2, s3):
"""
:type s1: str
:type s2: str
:type s3: str
:rtype: bool
"""
n1, n2, n3 = len(s1), len(s2), len(s3)
if n1 + n2 != n3:
return False
# dp[i][j] s1前i个字符串, s2前j个字符串， s3前i+j个字符串 是否能符合
dp = [[0] * (n2+1) for _ in xrange(n1+1)] # 先n2再n1 。。
for i in xrange(n1+1):
for j in xrange(n2+1):
if i == 0 and j == 0:
dp[0][0] = True # s1 empty s2 empty s3 empty
elif i == 0: # s1 is empty
dp[i][j] = dp[i][j-1] and s2[j-1] == s3[i+j-1]
elif j == 0: # s2 is empty
dp[i][j] = dp[i-1][j] and s1[i-1] == s3[i+j-1]
else:
dp[i][j] = (dp[i-1][j] and s1[i-1] == s3[i+j-1]) or (dp[i][j-1] and s2[j-1] == s3[i+j-1])
return dp[-1][-1]10. Regular Expression Matching
这题要再过一遍，不看思路的做
2.https://blog.csdn.net/hk2291976/article/details/51165010 图
class Solution(object):
def isMatch(self, s, p):
"""
:type s: str
:type p: str
:rtype: bool
"""
n1, n2 = len(s), len(p)
dp = [[False for x in xrange(n2+1)] for _ in xrange(n1+1)] # 加1 因为最开始是空和空匹配是True
dp[0][0] = True
for j in xrange(n2):
if p[j] == '*' and dp[0][j-1]: # 星号的 减两位检查 等于 a* = empty
dp[0][j+1] = True
for i in xrange(n1):
for j in xrange(n2):
if p[j] == '.':
dp[i+1][j+1] = dp[i][j] # i+1, j+1 是因为一开始是空和空匹配 i=0和j=0时候
elif p[j] == s[i]:
dp[i+1][j+1] = dp[i][j]
elif p[j] == '*':
if p[j-1] != s[i] and p[j-1] != '.':
dp[i+1][j+1] = dp[i+1][j-1]
else:
dp[i+1][j+1] = dp[i+1][j] or dp[i][j+1] or dp[i+1][j-1]
return dp[-1][-1]583. Delete Operation for Two Strings
经典题
Given two words word1 and word2, find the minimum number of steps required to make word1 and word2 the same, where in each step you can delete one character in either string.
Example 1:
Input: "sea", "eat"
Output: 2
Explanation: You need one step to make "sea" to "ea" and another step to make "eat" to "ea".
class Solution(object):
def minDistance(self, word1, word2):
"""
:type word1: str
:type word2: str
:rtype: int
"""
# len - length of max common str
l1, l2 = len(word1), len(word2)
dp = [[0]*(l2+1) for i in xrange(l1+1)] # find the common lenght
longest = 0
for i in xrange(l1):
for j in xrange(l2):
if word1[i] == word2[j]:
dp[i+1][j+1] = dp[i][j] + 1
else:
dp[i+1][j+1] = max(dp[i][j+1], dp[i+1][j])
return l1 + l2 - dp[l1][l2]*2