Leetcode - Product Of Array Except Self

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.)

[分析]
基本思路：数组中第 i 个元素除自身外的数组乘积=product(1,i-1) * product(i + 1, N -1)，
因此开辟两个数组prior 和 after，prior[i]=product(1,i-1), after[i]=product(i + 1, N -1)。
优化思路：在基本思路基础上如何优化空间效率呢，能否只使用常数的额外空间？注意到prior[i]=prior[i-1]*nums[i-1],且我们只需要当前的prior[i]，因此只使用一个变量即可，每次迭代时更新即可，而原来的after[]就可以直接存放在结果数组中，结果从前往后计算，计算到i位置时，i前面的都是最终结果，i后面的则是基本思路中after[]数组内容。

public class Solution {
    // base version
    public int[] productExceptSelf1(int[] nums) {
        if (nums == null) return null;
        int N = nums.length;
        int[] ret = new int[N];
        int[] prior = new int[N];
        int[] after = new int[N];
        prior[0] = 1;
        for (int i = 1; i < N; i++) {
            prior[i] = prior[i - 1] * nums[i - 1];
        }
        after[N - 1] = 1;
        for (int i = N - 2; i >= 0; i--) {
            after[i] = after[i + 1] * nums[i + 1];
        }
        for (int i = 0; i < N; i++) {
            ret[i] = prior[i] * after[i];
        }
        return ret;
    }
    // optimized version
    public int[] productExceptSelf(int[] nums) {
        if (nums == null) return null;
        int N = nums.length;
        int[] ret = new int[N];
        ret[N - 1] = 1;
        for (int i = N - 2; i >= 0; i--) {
            ret[i] = ret[i + 1] * nums[i + 1];
        }
        int prior = 1;
        for (int i = 0; i < N; i++) {
            ret[i] = prior * ret[i];
            prior *= nums[i];
        }
        return ret;
    }
}