LeetCode 238. Product of Array Except Self-优快云博客

本文介绍了一种高效算法，在O(n)时间内计算出一个数组中每个元素对应排除自身外其他所有元素的乘积结果，且整个过程不使用除法运算。通过两次遍历数组实现这一目标。

#include <iostream>
#include <vector>
using namespace std;

// Given an array of n integers where n > 1, nums, return an array ouput 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) time.
// for example, given [1, 2, 3, 4] -> [24, 12, 8, 6]

vector<int> productArray(vector<int> array) {
    // all we need to do is to go through the array twice.
    int size = array.size();
    vector<int> forward(size, 1);
    vector<int> backward(size, 1);
    for(int i = 0; i <= size - 2; ++i) {
       forward[i+1] = forward[i]* array[i];
    }

    for(int i = size - 1; i > 0; --i) {
        backward[i-1] = backward[i] * array[i];
    }

    for(int i = 0; i < size; ++i) {
        forward[i] = forward[i] * backward[i];
    }
    return forward;
}

// test
int main(void) {
    vector<int> testArray;
    testArray.push_back(1);
    testArray.push_back(2);
    testArray.push_back(3);
    testArray.push_back(4);
    vector<int> result = productArray(testArray);
    for(int i = 0; i < result.size(); ++i) {
        cout << result[i] << endl;
        cout << endl;
    }
}