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 ]

这里有一个小的递归函数(在c++中)来进行修改。它需要O(n)额外的空间(在堆栈上)。假设数组在a中，N表示数组长度，我们有:

int multiply(int *a, int fwdProduct, int indx) {
    int revProduct = 1;
    if (indx < N) {
       revProduct = multiply(a, fwdProduct*a[indx], indx+1);
       int cur = a[indx];
       a[indx] = fwdProduct * revProduct;
       revProduct *= cur;
    }
    return revProduct;
}

    int[] arr1 = { 1, 2, 3, 4, 5 };
    int[] product = new int[arr1.Length];              

    for (int i = 0; i < arr1.Length; i++)
    {
        for (int j = 0; j < product.Length; j++)
        {
            if (i != j)
            {
                product[j] = product[j] == 0 ? arr1[i] : product[j] * arr1[i];
            }
        }
    }

鬼鬼祟祟地绕过“不划分”规则:

sum = 0.0
for i in range(a):
  sum += log(a[i])

for i in range(a):
  output[i] = exp(sum - log(a[i]))

public static void main(String[] args) {
    int[] arr = { 1, 2, 3, 4, 5 };
    int[] result = { 1, 1, 1, 1, 1 };
    for (int i = 0; i < arr.length; i++) {
        for (int j = 0; j < i; j++) {
            result[i] *= arr[j];

        }
        for (int k = arr.length - 1; k > i; k--) {
            result[i] *= arr[k];
        }
    }
    for (int i : result) {
        System.out.println(i);
    }
}

我想出了这个解决方案，我发现它很清楚，你觉得呢!?

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);