LeetCode: 238. Product of Array Except Self

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.)

解题思路

  1. 如果数组中没有 0，则计算数组中所有数的乘积 P。除去第 i 位元素的乘积位 P/nums[i]
  2. 如果数组中有一个 0，则单独计算除开这一位的乘积。而其他位的 Product of Array Except Self 都为 0
  3. 如果数组中有两个及以上的 0，Product of Array Except Self 都为 0

AC 代码

class Solution {
public:
    vector<int> productExceptSelf(vector<int>& nums) {
        vector<int> products;
        
        long long int total_product = 1;
        int zero_count = 0;
        for(auto iter = nums.begin(); iter != nums.end(); ++iter)
        {
            if(*iter == 0)
            {
                ++zero_count;
            }
            else
            {
                total_product *= (*iter);
            }
        }
        
        for(auto iter = nums.begin(); iter != nums.end(); ++iter)
        {
            if(zero_count == 0)
            {
                products.push_back(total_product/(*iter));
            }
            else if(zero_count == 1 && *iter == 0)
            {
                products.push_back(total_product);
            }
            else
            {
                products.push_back(0);
            }
        }
        
        return products;
    }
};