Product of Array Except Self

最新推荐文章于 2022-10-27 17:29:32 发布

brucehb

最新推荐文章于 2022-10-27 17:29:32 发布

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本文介绍了一个解决数组乘积问题的方法，即返回数组中每个元素的乘积，但排除该元素自身。通过双重循环实现，确保不使用除法操作，并保持O(n)的时间复杂度。

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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 n = nums.size();

        vector<int> right(n+1, 1);
        for (int i = n-1; i > 0; i--)
        {
            right[i] = right[i+1]*nums[i];
        }

        int left = 1;
        vector<int> result;
        for (int i = 0; i < n; i++)
        {
            result.push_back(left*right[i+1]);
            left *= nums[i];
        }

        return result;
    }
};