伪代码:
A = [1,2,3,4,5,6,7,8,9]
B = [4,5,6,7,1,2,6,7,8]
count = 0
for i in range(len(A)):
for j in range(len(B)):
if A[i:i+3] == B[j:j+3]: #check 3 consecutive numbers if are equal
count += 1
print x[i:i+3]
print count
问题:我怎样才能实现 when A[4,5,6] == B[4,5,6]
,然后跳到A[6,7,8]==B[6,7,8]
,而不是A[5,6,7]==B[5,6,7]