题目：
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：
不能使用除法，那么就考虑分两步来乘，i之前的乘法preMultiply()，和i之后的乘法afterMultiply()。
Java
package nowcoder;
public class S51_Multiply {
public int[] multiply(int[] A){
int length = A.length;
int[] B = new int[length];
for (int i=0;i<length;i++){
B[i] = preMultiply(A, i) * afterMultiply(A, i);
}
return B;
}
private int preMultiply(int[] A, int a){
int pre = 1;
for (int i=0;i<a;i++)
pre *= A[i];
return pre;
}
private int afterMultiply(int[] A, int a){
int after = 1;
for (int i=A.length-1;i>a;i--)
after *= A[i];
return after;
}
public static void main(String[] args){
S51_Multiply s51 = new S51_Multiply();
int[] A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
int[] res = s51.multiply(A);
for (int i=0;i<res.length;i++)
System.out.print(res[i]+" ");
}
}
Java 思路二
package nowcoder;
public class S51_Multiply_new {
public int[] multiply(int[] A){
int length = A.length;
int[] B = new int[length];
for (int i=0;i<length;i++){
int temp = A[i];
B[i] = 1;
for (int j=0;j<length;j++){
A[i] = 1;
B[i] *= A[j];
}
A[i] = temp;//还原A[i],否则A[i]就是1了
}
return B;
}
public static void main(String[] args){
S51_Multiply_new s51 = new S51_Multiply_new();
int[] A = {1, 2, 3, 4};
int[] res = s51.multiply(A);
for (int i=0;i<res.length;i++)
System.out.print(res[i]+" ");
}
}
Python
class Multiply:
def multiply(self, A):
length = len(A)
B = []
for i in range(length):
temp = A[i]
b = 1
for j in range(length):
A[i] = 1
b *= A[j]
B.append(b)
A[i] = temp
return B
if __name__ == '__main__':
test = Multiply()
A = [1, 2, 3, 4, 5]
print(test.multiply(A))
