构建乘积数组

思路：

B[i]=A[0]A[1]...A[i-1]A[i+1]...A[n-1]，求A数组的连乘，但不包含A[i];

1. 如果暴力，O(n²)
2. 递归解决，分为两部分：
   • 计算前部分： B[0] = 1
     B[1] = A[1-1] * B[0] = A[0]
     B[2] = A[2-1] * B[1] = A[0]* A[1]
     ...
     B[i] = A[i-1] * B[i-1]... = A[0] A[1] ...A[i-1]
   • 计算后部分：i为最后一个索引下标
     B[i] = 1* B[i] = A[0]* A[1] * ... * A[i-1]
     B[i-1] = B[i-1] * A[i] = A[0] * A[1] ... A[i-2] * A[i]
     ...
     B[0] = B[0] * A[1] * ... * A[i-1] = A[1] * ...* A[i-2] *A[i]

后半部分

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

代码

/**
 * B[i]=A[0]*A[1]*...*A[i-1]*A[i+1]*...*A[n-1]
 * Created by liqiushi on 2018/3/24.
 */
public class Multiply {
    public int[] multiply(int[] A) {
        if (A == null) {
            return null;
        }
        int[] B = new int[A.length];
        if (B.length == 0) {
            return B;
        }
        B[0] = 1;
        //先计算了 B[i] = A[0]*A[1]*A[2]*...*A[i-1]
        for (int i = 1; i < A.length; i++) {
            B[i] = A[i - 1] * B[i - 1];
        }
        //后部分:
        int tmp = 1;
/*        for (int i = A.length - 2; i >= 0; i--) {
            tmp *= A[i+1];
            B[i] = tmp * B[i];
        }*/
        for (int i = A.length - 1; i >= 0; i--) {
            B[i] = tmp * B[i];
            tmp *= A[i];
        }
        
        return B;
    }
}