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;
};