最长公共子序列的python实现

def subsequence():
	m = len(A)
	n = len(B)
	log_lengths = [[0 for j in range(m)] for i in range(n)]
	for i in range(n):
		for j in range(m):
			if B[i] == A[j]:
				if i>0 and j>0:
					log_lengths[i][j] = log_lengths[i-1][j-1]+1
				else:
					log_lengths[i][j] = 1
			else:
				if i>0 and j>0:
					log_lengths[i][j] = max(log_lengths[i-1][j],log_lengths[i][j-1])
				elif i>0 and j==0:
					log_lengths[i][j] = log_lengths[i-1][j]
				elif i==0 and j>0:
					log_lengths[i][j] = log_lengths[i][j-1]
	maxs = max(map(max,log_lengths))
	print("the subsequence length is :%d"%maxs)

这里的空间复杂度是o(n^2)，其实是可以优化到o(n)的，具体思路可以看我这篇文章