238. Product of Array Except Self

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.)

本來想說要用2個arrays加上DP的想法做，但後來想想可以只需要兩個額外的left和right變數紀錄當前掃描到的element的左和右的乘積各是多少。

vector<int> productExceptSelf(vector<int>& nums) { // time: O(n); space: O(1)
    int n = nums.size(), left = 1, right = 1;
    vector<int> res(n, 1);
    for (int i = 0; i < n; ++i) {
        res[i] *= left;
        left *= nums[i];
    }
    for (int j = n - 1; j >= 0; --j) {
        res[j] *= right;
        right *= nums[j];
    }
    return res;
}

可以把左到右、右到左的兩個迴圈合併成一個。

vector<int> productExceptSelf(vector<int>& nums) { // time: O(n); space: O(1)
    int n = nums.size(), left = 1, right = 1;
    vector<int> res(n, 1);
    for (int i = 0; i < n; ++i) {
        res[i] *= left;
        left *= nums[i];
        res[n - i - 1] *= right;
        right *=  nums[n - i - 1];
    }
    return res;
}