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LCS算法的python实现

2018年02月15日 ⁄ 综合 ⁄ 共 941字 ⁄ 字号 评论关闭
'''
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)

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