238 - Product of Array Except Self

不使用除法计算乘积

最新推荐文章于 2020-11-24 22:15:29 发布

原创最新推荐文章于 2020-11-24 22:15:29 发布 · 468 阅读

0 ·

CC 4.0 BY-SA版权

LeetCode 专栏收录该内容

18 篇文章

订阅专栏

本文介绍了一种算法，可以在O(n)的时间复杂度内计算出数组中每个元素对应的除自身之外所有元素的乘积，并提供了一个C++实现示例。

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.)

C++ 代码如下（涵盖测试代码）：

#include <iostream>
#include <vector>

using namespace std;

class Solution {
public:
    vector<int> productExceptSelf(vector<int>& nums) {
        if(nums.size() <= 0)
                return nums;
        int products = 1;
        vector<int> zeroNum;
        for(int i = 0; i < nums.size(); i++)
                if(nums[i] != 0)
                        products *=  nums[i];
                else {
                        zeroNum.push_back(i);
                }

        //cout << "output" << endl; 
        vector<int> result(nums.size(), 0);
        if(zeroNum.size() >= 2) {
                return result;
        }

        if(zeroNum.size() == 1) {
                result[zeroNum[0]] = products;
                return result;
        }
        for(int i = 0; i < nums.size(); i++) {
                result[i] = products/nums[i] ;
        }
        return result;
    }
};

int main() {
        vector<int> nums;
        nums.push_back(0);
        nums.push_back(2);
        nums.push_back(0);
        nums.push_back(4);

        Solution s;
        vector<int> r = s.productExceptSelf(nums);
        for(int i = 0; i < r.size(); i++)
                cout << r[i] << endl;
}

注：提交两遍才过，第一遍忘记了考虑整个数组中有 0 的情况，加上判断，就通过了。

思路很简单，关键是细节。