51 构建乘积数组

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：先计算下三角形每一行的乘积。
B[0] = 1
B[1] = A0
B[2] = A0A1
B[3] = A0A1A2
…
B[n-1] = A0A1*A2…*An-2
代码实现如下：

        B[0] = 1;
        for (int i = 1; i < A.length; i++) {
            B[i] = B[i - 1] * A[i - 1];
        }

然后在上面步骤的基础上乘以上三角形：
B[n-2] = B[n-2] * An-1
B[n-3] = B[n-3] * An-2*An-1
…
B[1] = B1 * A2 * … * An-1
B[0] = B0 * A1 * A2 * … * An-1
代码实现如下：

        int temp = 1;
        for (int j = A.length - 2; j >= 0; j--) {
            temp *= A[j + 1];
            B[j] *= temp;
        }

第二种方法，双层for循环，遇到A[i]就continue

public class Test51 {
    public static int[] multiply(int[] A) {
        int[] B = new int[A.length];
        for (int i = 0; i < B.length; i++) {
            B[i] = 1;
            for (int j = 0; j < A.length; j++) {
                if (j == i) {
                    continue;
                }
                B[i] *= A[j];
            }
        }
        return B;
    }

    public static int[] multiply2(int[] A) {
        int[] B = new int[A.length];
        B[0] = 1;
        for (int i = 1; i < A.length; i++) {
            B[i] = B[i - 1] * A[i - 1];
        }
        int temp = 1;
        for (int j = A.length - 2; j >= 0; j--) {
            temp *= A[j + 1];
            B[j] *= temp;
        }
        return B;
    }

    public static void main(String[] args) {
        int[] a = new int[]{5, 1, 2, 3};
        int[] b = new int[]{5, 1, 2, 3};
        int[] c = multiply(a);
        int[] d = multiply(b);
        for (int i = 0; i < c.length; i++) {
            System.out.print(c[i] + " ");
        }
        System.out.println();
        for (int i = 0; i < c.length; i++) {
            System.out.print(d[i] + " ");
        }
    }
}