我在一次工作面试中被问到这个问题,我想知道其他人是如何解决这个问题的。我最擅长使用Java,但也欢迎使用其他语言的解决方案。

给定一个数字数组nums,返回一个数字数组products,其中products[i]是所有nums[j]的乘积,j != i。 输入:[1,2,3,4,5] 输出:[(2 * 3 * 4 * 5),(1 * 3 * 4 * 5),(1 * 2 * 4 * 5),(1 * 2 * 3 * 5),(1 * 2 * 3 * 4)] = [120, 60, 40, 30, 24] 你必须在O(N)中不使用除法来做这个。


当前回答

int[] b = new int[] { 1, 2, 3, 4, 5 };            
int j;
for(int i=0;i<b.Length;i++)
{
  int prod = 1;
  int s = b[i];
  for(j=i;j<b.Length-1;j++)
  {
    prod = prod * b[j + 1];
  }
int pos = i;    
while(pos!=-1)
{
  pos--;
  if(pos!=-1)
     prod = prod * b[pos];                    
}
Console.WriteLine("\n Output is {0}",prod);
}

其他回答

这是我的代码:

int multiply(int a[],int n,int nextproduct,int i)
{
    int prevproduct=1;
    if(i>=n)
        return prevproduct;
    prevproduct=multiply(a,n,nextproduct*a[i],i+1);
    printf(" i=%d > %d\n",i,prevproduct*nextproduct);
    return prevproduct*a[i];
}

int main()
{
    int a[]={2,4,1,3,5};
    multiply(a,5,1,0);
    return 0;
}

这是ptyhon版本

  # This solution use O(n) time and O(n) space
  def productExceptSelf(self, nums):
    """
    :type nums: List[int]
    :rtype: List[int]
    """
    N = len(nums)
    if N == 0: return

    # Initialzie list of 1, size N
    l_prods, r_prods = [1]*N, [1]*N

    for i in range(1, N):
      l_prods[i] = l_prods[i-1] * nums[i-1]

    for i in reversed(range(N-1)):
      r_prods[i] = r_prods[i+1] * nums[i+1]

    result = [x*y for x,y in zip(l_prods,r_prods)]
    return result

  # This solution use O(n) time and O(1) space
  def productExceptSelfSpaceOptimized(self, nums):
    """
    :type nums: List[int]
    :rtype: List[int]
    """
    N = len(nums)
    if N == 0: return

    # Initialzie list of 1, size N
    result = [1]*N

    for i in range(1, N):
      result[i] = result[i-1] * nums[i-1]

    r_prod = 1
    for i in reversed(range(N)):
      result[i] *= r_prod
      r_prod *= nums[i]

    return result

上下两次。在O(N)完成的工作

private static int[] multiply(int[] numbers) {
        int[] multiplied = new int[numbers.length];
        int total = 1;

        multiplied[0] = 1;
        for (int i = 1; i < numbers.length; i++) {
            multiplied[i] = numbers[i - 1] * multiplied[i - 1];
        }

        for (int j = numbers.length - 2; j >= 0; j--) {
            total *= numbers[j + 1];
            multiplied[j] = total * multiplied[j];
        }

        return multiplied;
    }
int[] b = new int[] { 1, 2, 3, 4, 5 };            
int j;
for(int i=0;i<b.Length;i++)
{
  int prod = 1;
  int s = b[i];
  for(j=i;j<b.Length-1;j++)
  {
    prod = prod * b[j + 1];
  }
int pos = i;    
while(pos!=-1)
{
  pos--;
  if(pos!=-1)
     prod = prod * b[pos];                    
}
Console.WriteLine("\n Output is {0}",prod);
}

我用Javascript想出了两个解决方案,一个有除法,一个没有

//不除法 函数methodOne(arr) { 加勒比海盗。Map (item => { 加勒比海盗。Reduce ((result, num) => { If (num !== item) { 结果=结果* num; } 返回结果; }, 1) }); } //使用除法 函数methodTwo(arr) { Var mul = arr。Reduce ((result, num) => { 结果=结果* num; 返回结果; }, 1) 加勒比海盗。Map (item => mul/item); } console.log(methodOne([1,2,3,4,5])); console.log(methodTwo([1,2,3,4,5]));