Product of Array Except Self Leetcode

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.)

 

 

 

这道题不会做了。。。但其实可以把问题分解一下，除了本身之外所有的数的乘积可以分为，这个数前面的所有的数的乘积乘上这个数后面所有的数的乘积，所有问题就迎刃而解啦。

下面是优化过的没有用extra space的代码，其实用了extra的代码更好理解，怪不得是follow up.

public class Solution {
    public int[] productExceptSelf(int[] nums) {
        int[] output = new int[nums.length];
        output[0] = 1;
        int product = 1;
        for (int i = 1; i < nums.length; i++) {
            output[i] = nums[i - 1] * output[i - 1];
        }
        for (int i = nums.length - 2; i >= 0; i--) {
            output[i] = nums[i + 1] * product * output[i];
            product *= nums[i + 1];
        }
        return output;
    }
}