Leetcode 238. 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.)

求一个数组除去每一个位置以后的乘积。不能用除法，O（n）时间复杂度，O(1)空间复杂度。

两个量分别统计前缀数组乘积和后缀数组乘积，每一个位置的值为前缀和后缀的乘积相乘。

class Solution { public: vector<int> productExceptSelf(vector<int>& nums) { vector<int> res(nums.size(), 1); int s = 1, e = 1; int n = nums.size(); for(int i = 0; i < nums.size(); i++) { res[i] *= s; s *= nums[i]; res[n - 1 - i] *= e; e *= nums[n - 1 - i]; } return res; } };