QUESTION:

Given an array of n integers where n > 1, nums, return an array output such that output[i] is equal to the product of all the elements of nums except nums[i].

Solve it without division and in O(n).

For example, given [1,2,3,4], return [24,12,8,6].

Follow up: Could you solve it with constant space complexity? (Note: The output array does not count as extra space for the purpose of space complexity analysis.)

EXPLANATION:

这道题目的迷惑的地方就是在于如何构造对应的数组了。如果可以采用对应的除法的话应该是很容易就可以搞定的，但是题目规定了不能够使用对应的除法。那么我们就可以这样构造这个数组。

a = { 1， a[0], a[0]*a[1], a[0]*a[1]*a[2]}

b = { a[1]*a[2]*a[3] , a[2]*a[3] , a[3] , 1 }

这两个数组相乘，就可以得到对应的值了，比如 result[0] = a[0]*b[0]

那么这两个数组如何构建其实就很容易了。

SOLUTION:

public class Solution {
    public int[] productExceptSelf(int[] nums) {
        int[] result = new int[nums.length];
        result[0] = 1;
        for(int i = 1;i<nums.length;i++){
            result[i] = nums[i-1]*result[i-1];
        }
        int right = 1;
        for(int i = nums.length-1;i>=0;i--){
            result[i]*=right;
            right*=nums[i];
        }
        return result;   
    }
}