Difficulty: Medium
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.)
class Solution { public: vector<int> productExceptSelf(vector<int>& nums) { int s = nums.size(), i = 0; vector<int> <code>output</code>(s, 1); for (i = 1; i < s; ++i) <code>output</code>[i] = <code>output</code>[i-1] * nums[i-1]; int k = nums[s-1]; for (i = s - 2; i >=0; --i) { <code>output</code>[i] *= k; k *= nums[i]; } return <code>output</code>; } }; 思路是先计算 output[i]左边所有数的乘积，再计算 output[i]右边所有数的乘积，用两个for循环来完成计算。
