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.)

解：
这几道题都没想出来啊。。
基本思想就是先遍历一遍，用数组存储前i个元素的乘积但是往右边挫了一个，也就是不包含当前i的前i-1个元素的积，然后再从右边扫一遍，记录后几个元素乘积

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