[leetcode] Product of Array Except Self

最新推荐文章于 2024-10-03 07:19:37 发布

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本文提供了一种方法来解决给定数组中每个元素的乘积问题，无需使用除法运算，并且算法的时间复杂度保持在O(n)。同时探讨了进阶挑战，即在常数空间复杂度下解决问题。

From : https://leetcode.com/problems/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.)

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Solution :

class Solution {
public:
    vector<int> productExceptSelf(vector<int>& nums) {
        int n=nums.size(), right=nums[n-1];
        vector<int> ans(n, 1);
        for(int i=1; i<n; i++) {
            ans[i] = nums[i-1]*ans[i-1];
        }
        for(int i=n-2; i>=0; i--) {
            ans[i] = right*ans[i];
            right *= nums[i];
        }
        return ans;
    }
};