lintcode-easy-Product of Array Exclude Itself

Posted 2020-06-17 哥布林工程师

Given an integers array A.

Define B[i] = A[0] * ... * A[i-1] * A[i+1] * ... * A[n-1], calculate B WITHOUT divide operation.

For A = [1, 2, 3], return [6, 3, 2].

public class Solution {
    /**
     * @param A: Given an integers array A
     * @return: A Long array B and B[i]= A[0] * ... * A[i-1] * A[i+1] * ... * A[n-1]
     */
    public ArrayList<Long> productExcludeItself(ArrayList<Integer> A) {
        // write your code
        ArrayList<Long> result = new ArrayList<Long>();
        if(A == null || A.size() == 0)
            return result;
        
        long[] left2curr = new long[A.size()];
        long[] curr2right = new long[A.size()];
        
        left2curr[0] = 1;
        curr2right[A.size() - 1] = 1;
        
        for(int i = 1; i < A.size(); i++)
            left2curr[i] = A.get(i - 1) * left2curr[i - 1];
        for(int i = A.size() - 2; i >= 0; i--)
            curr2right[i] = A.get(i + 1) * curr2right[i + 1];
        
        for(int i = 0; i < A.size(); i++)
            result.add(left2curr[i] * curr2right[i]);
        
        return result;
    }
}