LCS算法的python实现

'''
Created on 2012-11-9

@author: Pandara
'''
def lcs_len(a, b):
'''
a, b: strings
'''
n = len(a)
m = len(b)

l = [([0] * (m + 1)) for i in range(n + 1)]
direct = [([0] * m) for i in range(n)]#0 for top left, -1 for left, 1 for top

for i in range(n + 1)[1:]:
for j in range(m + 1)[1:]:
if a[i - 1] == b[j - 1]:
l[i][j] = l[i - 1][j - 1] + 1
elif l[i][j - 1] > l[i - 1][j]:
l[i][j] = l[i][j - 1]
direct[i - 1][j - 1] = -1
else:
l[i][j] = l[i - 1][j]
direct[i - 1][j - 1] = 1

return l, direct

def get_lcs(direct, a, i, j):
'''
direct: martix of arrows
a: the string regarded as row
i: len(a) - 1, for initialization
j: len(b) - 1, for initialization
'''
lcs = []
get_lcs_inner(direct, a, i, j, lcs)
return lcs

def get_lcs_inner(direct, a, i, j, lcs):
if i < 0 or j < 0:
return

if direct[i][j] == 0:
get_lcs_inner(direct, a, i - 1, j - 1, lcs)
lcs.append(a[i])

elif direct[i][j] == 1:
get_lcs_inner(direct, a, i - 1, j, lcs)
else:
get_lcs_inner(direct, a, i, j - 1, lcs)

if __name__ == "__main__":
a = "abcbdab"
b = "bdcaba"

l, direct = lcs_len(a, b)
lcs = get_lcs(direct, a, len(a) - 1, len(b) - 1)

print "the length of lcs is:", l[len(a)][len(b)]
print "one of the lcs:", "".join(lcs)