给定一个数组A[0,1,...,n-1],请构建一个数组B[0,1,...,n-1],其中B中的元素B[i]=A[0]*A[1]*...*A[i-1]*A[i+1]*...*A[n-1]。不能使用除法。
思路:
B[0] | 1 | A[1] | A[2] | ..... | A[n-3] | A[n-2] | A[n-1] |
B[1] | A[0] | 1 | A[2] | ..... | A[n-3] | A[n-2] | A[n-1] |
B[2] | A[0] | A[1] | 1 | A[2] | ....... | A[n-2] | A[n-1] |
B[3] | A[0] | A[1] | A[2] | 1 | A[4] | ...... | A[n-1] |
B[4] | A[0] | A[1] | A[2] | A[3] | 1 | A[5] | ...... |
B[i]=下三角乘以上三角
下三角=b[i-1]*A[i-1]
上三角=1*A[n-1]*...*A[i+!]
def multiply(self, A):
# write code here
if A==[]:
return []
B=[1]*len(A)
for i in range(1,len(A)):
B[i]=B[i-1]*A[i-1]
ans=1
for j in range(len(A)-2,-1,-1):
ans*=A[j+1]
B[j]*=ans
return B