238 - Product of Array Except Self

最新推荐文章于 2020-11-24 22:15:29 发布

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本文介绍了一种高效算法，用于求解给定数组中每个元素对应的除自身以外所有元素的乘积，且不使用除法操作。该算法遵循O(n)的时间复杂度要求，并探讨了常数空间复杂度的解决方案。

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Problem

Given an array nums of n integers where n > 1, return an array output such that output[i] is equal to the product of all the elements of nums except nums[i].

Example:

Input:  [1,2,3,4]
Output: [24,12,8,6]

Note: Please solve it without division and in O(n).

Follow up:
Could you solve it with constant space complexity? (The output array does not count as extra space for the purpose of space complexity analysis.)

Code

class Solution {
public:
    vector<int> productExceptSelf(vector<int>& nums) {
        vector<int> vec; 
        int size = nums.size(); 
        if (size == 0 || size == 1) {
            return vec;
        }
        int t = 1, a[size], c[size];
        for (int i = 0; i < size; ++i) {
            a[i] = t * nums[i];
            t *= nums[i];
        }
        t = 1;
        for (int i = size - 1; i >= 0; --i) {
            c[i] = t * nums[i];
            t = t * nums[i];
        }
        vec.push_back(c[1]);
        for (int i = 1; i < size - 1; ++i) {
            vec.push_back(a[i - 1] * c[i + 1]);
        }
        vec.push_back(a[size - 2]);
        return vec;
    }
};