題目來(lái)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.)
這道題看到的第一想法就是先把所有相乘，然后再除以每個(gè)nums[i]，然后還得處理一些特殊情況。

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

感覺(jué)代碼寫的比較亂…然后我還是想的太多，直接乘不好嗎對(duì)吧…代碼如下：

class Solution {
public:
    vector<int> productExceptSelf(vector<int>& nums) {
        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 a = 1;
        for (int i=n-1; i>0; i--) {
            a *= nums[i];
            res[i-1] *= a;
        }
        return res;
    }
};