Product of Array Except Self

题目来源
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.)
这道题看到的第一想法就是先把所有相乘，然后再除以每个nums[i]，然后还得处理一些特殊情况。

class Solution {
public:
    vector productExceptSelf(vector& nums) {
        long long sum = 1;
        vector res;
        int count0 = 0;
        for (auto num : nums) {
            if (num == 0) {
                count0++;
                continue;
            }
            sum *= num;
        }
        if (count0 > 1)
            return vector(nums.size(), 0);
        for (auto num : nums) {
            if (count0 == 0)
                res.push_back(sum / num);
            else {
                if (num == 0)
                    res.push_back(sum);
                else
                    res.push_back(0);
            }
        }
        return res;
    }
};

感觉代码写的比较乱…然后我还是想的太多，直接乘不好吗对吧…代码如下：

class Solution {
public:
    vector productExceptSelf(vector& nums) {
        int n = nums.size();
        vector res(n, 1);
        for (int i=1; i0; i--) {
            a *= nums[i];
            res[i-1] *= a;
        }
        return res;
    }
};