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minimum-ascii-delete-sum-for-two-strings.py 1.44 KB
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Allen Liu 提交于 2018-10-13 01:56 +08:00 . add complexity
# Time: O(m * n)
# Space: O(n)
class Solution(object):
def minimumDeleteSum(self, s1, s2):
"""
:type s1: str
:type s2: str
:rtype: int
"""
dp = [[0] * (len(s2)+1) for _ in xrange(2)]
for j in xrange(len(s2)):
dp[0][j+1] = dp[0][j] + ord(s2[j])
for i in xrange(len(s1)):
dp[(i+1)%2][0] = dp[i%2][0] + ord(s1[i])
for j in xrange(len(s2)):
if s1[i] == s2[j]:
dp[(i+1)%2][j+1] = dp[i%2][j]
else:
dp[(i+1)%2][j+1] = min(dp[i%2][j+1] + ord(s1[i]), \
dp[(i+1)%2][j] + ord(s2[j]))
return dp[len(s1)%2][-1]
# Time: O(m * n)
# Space: O(m * n)
class Solution2(object):
def minimumDeleteSum(self, s1, s2):
"""
:type s1: str
:type s2: str
:rtype: int
"""
dp = [[0] * (len(s2)+1) for _ in xrange(len(s1)+1)]
for i in xrange(len(s1)):
dp[i+1][0] = dp[i][0] + ord(s1[i])
for j in xrange(len(s2)):
dp[0][j+1] = dp[0][j] + ord(s2[j])
for i in xrange(len(s1)):
for j in xrange(len(s2)):
if s1[i] == s2[j]:
dp[i+1][j+1] = dp[i][j]
else:
dp[i+1][j+1] = min(dp[i][j+1] + ord(s1[i]), \
dp[i+1][j] + ord(s2[j]))
return dp[-1][-1]
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