【LeetCode】- Product of Array Except Self（除了自己的其它的数相乘）

1、题目描述

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.)

2、问题描述：

• 一个数组，将其它位上的元素之和作为本位上的值。

3、问题关键：

• 用一个数组，从左到右，先记录第i个元素左边元素的乘积，再从右到 左乘上右边元素的乘积。

4、C++代码：

class Solution {
public:
    vector productExceptSelf(vector& nums) {
        vector res(nums.size());
        for (int i = 0, t = 1;i < nums.size(); i ++) {
            res[i] = t;//记录左边元素的乘积。
            t *= nums[i];
        }
        for (int i = nums.size() - 1, t = 1; i >= 0; i --) {
            res[i] *= t;//再乘上右边元素的乘积。
            t *= nums[i];
        }
        return res;
    }
};