php版本 使用不除法的array_product函数。 如果我们将i的值临时设为1，那么数组product将完全满足我们的需要

<?php
function product($key, $arr)
{
    $arr[$key] = 1;
    return array_product($arr);
};
$arr = [1, 2, 3, 4, 5];
$newarr = array();


foreach ($arr as $key => $value) {

    $newarr[$key] = product($key, $arr);
}
print_r($newarr);

我们可以先从列表中排除nums[j](其中j != i)，然后得到其余部分的乘积;下面是python解决这个难题的方法:

from functools import reduce
def products(nums):
    return [ reduce(lambda x,y: x * y, nums[:i] + nums[i+1:]) for i in range(len(nums)) ]
print(products([1, 2, 3, 4, 5]))

[out]
[120, 60, 40, 30, 24]

import java.util.Arrays;

public class Pratik
{
    public static void main(String[] args)
    {
        int[] array = {2, 3, 4, 5, 6};      //  OUTPUT: 360  240  180  144  120
        int[] products = new int[array.length];
        arrayProduct(array, products);
        System.out.println(Arrays.toString(products));
    }

    public static void arrayProduct(int array[], int products[])
    {
        double sum = 0, EPSILON = 1e-9;

        for(int i = 0; i < array.length; i++)
            sum += Math.log(array[i]);

        for(int i = 0; i < array.length; i++)
            products[i] = (int) (EPSILON + Math.exp(sum - Math.log(array[i])));
    }
}

输出:

[360, 240, 180, 144, 120]

时间复杂度:O(n) 空间复杂度:O(1)

下面是我用现代c++编写的解决方案。它使用std::transform，很容易记住。

在线代码(wandbox)。

#include<algorithm>
#include<iostream>
#include<vector>

using namespace std;

vector<int>& multiply_up(vector<int>& v){
    v.insert(v.begin(),1);
    transform(v.begin()+1, v.end()
             ,v.begin()
             ,v.begin()+1
             ,[](auto const& a, auto const& b) { return b*a; }
             );
    v.pop_back();
    return v;
}

int main() {
    vector<int> v = {1,2,3,4,5};
    auto vr = v;

    reverse(vr.begin(),vr.end());
    multiply_up(v);
    multiply_up(vr);
    reverse(vr.begin(),vr.end());

    transform(v.begin(),v.end()
             ,vr.begin()
             ,v.begin()
             ,[](auto const& a, auto const& b) { return b*a; }
             );

    for(auto& i: v) cout << i << " "; 
}

使用EcmaScript 2015编码

'use strict'

/*
Write a function that, given an array of n integers, returns an array of all possible products using exactly (n - 1) of those integers.
*/
/*
Correct behavior:
- the output array will have the same length as the input array, ie. one result array for each skipped element
- to compare result arrays properly, the arrays need to be sorted
- if array lemgth is zero, result is empty array
- if array length is 1, result is a single-element array of 1

input array: [1, 2, 3]
1*2 = 2
1*3 = 3
2*3 = 6
result: [2, 3, 6]
*/
class Test {
  setInput(i) {
    this.input = i
    return this
  }
  setExpected(e) {
    this.expected = e.sort()
    return this
  }
}

class FunctionTester {
  constructor() {
    this.tests = [
      new Test().setInput([1, 2, 3]).setExpected([6, 3, 2]),
      new Test().setInput([2, 3, 4, 5, 6]).setExpected([3 * 4 * 5 * 6, 2 * 4 * 5 * 6, 2 * 3 * 5 * 6, 2 * 3 * 4 * 6, 2 * 3 * 4 * 5]),
    ]
  }

  test(f) {
    console.log('function:', f.name)
    this.tests.forEach((test, index) => {
      var heading = 'Test #' + index + ':'
      var actual = f(test.input)
      var failure = this._check(actual, test)

      if (!failure) console.log(heading, 'input:', test.input, 'output:', actual)
      else console.error(heading, failure)

      return !failure
    })
  }

  testChain(f) {
    this.test(f)
    return this
  }

  _check(actual, test) {
      if (!Array.isArray(actual)) return 'BAD: actual not array'
      if (actual.length !== test.expected.length) return 'BAD: actual length is ' + actual.length + ' expected: ' + test.expected.length
      if (!actual.every(this._isNumber)) return 'BAD: some actual values are not of type number'
      if (!actual.sort().every(isSame)) return 'BAD: arrays not the same: [' + actual.join(', ') + '] and [' + test.expected.join(', ') + ']'

      function isSame(value, index) {
        return value === test.expected[index]
      }
  }

  _isNumber(v) {
    return typeof v === 'number'
  }
}

/*
Efficient: use two iterations of an aggregate product
We need two iterations, because one aggregate goes from last-to-first
The first iteration populates the array with products of indices higher than the skipped index
The second iteration calculates products of indices lower than the skipped index and multiplies the two aggregates

input array:
1 2 3
   2*3
1*    3
1*2

input array:
2 3 4 5 6
    (3 * 4 * 5 * 6)
(2) *     4 * 5 * 6
(2 * 3) *     5 * 6
(2 * 3 * 4) *     (6)
(2 * 3 * 4 * 5)

big O: (n - 2) + (n - 2)+ (n - 2) = 3n - 6 => o(3n)
*/
function multiplier2(ns) {
  var result = []

  if (ns.length > 1) {
    var lastIndex = ns.length - 1
    var aggregate

    // for the first iteration, there is nothing to do for the last element
    var index = lastIndex
    for (var i = 0; i < lastIndex; i++) {
      if (!i) aggregate = ns[index]
      else aggregate *= ns[index]
      result[--index] = aggregate
    }

    // for second iteration, there is nothing to do for element 0
    // aggregate does not require multiplication for element 1
    // no multiplication is required for the last element
    for (var i = 1; i <= lastIndex; i++) {
      if (i === 1) aggregate = ns[0]
      else aggregate *= ns[i - 1]
      if (i !== lastIndex) result[i] *= aggregate
      else result[i] = aggregate
    }
  } else if (ns.length === 1) result[0] = 1

  return result
}

/*
Create the list of products by iterating over the input array

the for loop is iterated once for each input element: that is n
for every n, we make (n - 1) multiplications, that becomes n (n-1)
O(n^2)
*/
function multiplier(ns) {
  var result = []

  for (var i = 0; i < ns.length; i++) {
    result.push(ns.reduce((reduce, value, index) =>
      !i && index === 1 ? value // edge case: we should skip element 0 and it's the first invocation: ignore reduce
      : index !== i ? reduce * value // multiply if it is not the element that should be skipped
      : reduce))
  }

  return result
}

/*
Multiply by clone the array and remove one of the integers

O(n^2) and expensive array manipulation
*/
function multiplier0(ns) {
  var result = []

  for (var i = 0; i < ns.length; i++) {
    var ns1 = ns.slice() // clone ns array
    ns1.splice(i, 1) // remove element i
    result.push(ns1.reduce((reduce, value) => reduce * value))
  }

  return result
}

new FunctionTester().testChain(multiplier0).testChain(multiplier).testChain(multiplier2)

使用Node.js v4.4.5运行:

Node—harmony integerarrays.js

function: multiplier0
Test #0: input: [ 1, 2, 3 ] output: [ 2, 3, 6 ]
Test #1: input: [ 2, 3, 4, 5, 6 ] output: [ 120, 144, 180, 240, 360 ]
function: multiplier
Test #0: input: [ 1, 2, 3 ] output: [ 2, 3, 6 ]
Test #1: input: [ 2, 3, 4, 5, 6 ] output: [ 120, 144, 180, 240, 360 ]
function: multiplier2
Test #0: input: [ 1, 2, 3 ] output: [ 2, 3, 6 ]
Test #1: input: [ 2, 3, 4, 5, 6 ] output: [ 120, 144, 180, 240, 360 ]

我习惯使用c#:

    public int[] ProductExceptSelf(int[] nums)
    {
        int[] returnArray = new int[nums.Length];
        List<int> auxList = new List<int>();
        int multTotal = 0;

        // If no zeros are contained in the array you only have to calculate it once
        if(!nums.Contains(0))
        {
            multTotal = nums.ToList().Aggregate((a, b) => a * b);

            for (int i = 0; i < nums.Length; i++)
            {
                returnArray[i] = multTotal / nums[i];
            }
        }
        else
        {
            for (int i = 0; i < nums.Length; i++)
            {
                auxList = nums.ToList();
                auxList.RemoveAt(i);
                if (!auxList.Contains(0))
                {
                    returnArray[i] = auxList.Aggregate((a, b) => a * b);
                }
                else
                {
                    returnArray[i] = 0;
                }
            }
        }            

        return returnArray;
    }