238.Product of Array Except Self

Tags: Medium Array

Links: https://leetcode.com/problems/product-of-array-except-self/

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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]

Constraint: It's guaranteed that the product of the elements of any prefix or suffix of the array (including the whole array) fits in a 32 bit integer.

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.)

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class Solution {
public:
    vector<int> productExceptSelf(vector<int>& nums) {
        std::ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        cout.tie(NULL);

        int n = nums.size();
        vector<int> res(n, 1);

        for (int i = 1; i < n; ++i) {
            res[i] = res[i - 1] * nums[i - 1];
        }
        int tmp = 1;
        for (int i = n - 2; i >= 0; --i) {
            tmp *= nums[i + 1];
            res[i] *= tmp;
        }

        return res;
    }
};

题意是计算数组内所有数字的乘积，但是不包含当前位置的数字。一个想法是使用前缀乘积和后缀乘积，两个数组，也就是维护当前数字的左半部分的连乘结果和右半部分的连乘结果。这样就需要额外数组。

进一步优化，可以直接用结果数组的空间，先遍历一遍计算前缀乘积，然后逆序遍历的时候，用一个临时变量tmp保存后缀乘积的结果。