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.)
给定一个n个整数的数组,其中n> 1,nums返回一个数组输出,使得输出[i]等于除nums [i]之外的所有num的元素的乘积。 解决它没有分裂和O(n)。 例如,给出[1,2,3,4],返回[24,12,8,6]。
/*** @param {number[]} nums* @return {number[]}*/var productExceptSelf = function(nums) {let res = nums.concat().fill(1);for (let i = 1; i < nums.length; i++) {res[i] = res[i - 1] * nums[i - 1];}let right = 1;for (let i = nums.length - 1; i >= 0; i--) {res[i] *= right;right *= nums[i];}return res;};