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
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;
}
其他回答
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]))
我们正在分解数组的元素，首先从下标之前开始，即前缀，然后是下标或后缀之后
class Solution:
def productExceptSelf(nums):
length = len(nums)
result = [1] * length
prefix_product = 1
postfix_product = 1
# we initialize the result and products
for i in range(length)
result[i] *= prefix_product
prefix_product *= nums[i]
#we multiply the result by each number before the index
for i in range(length-1,-1,-1)
result[i] *= postfix_product
postfix_product *= nums[i]
#same for after index
return result
抱歉，走路时用手机
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];
}
}
}
试试这个!
import java.util.*;
class arrProduct
{
public static void main(String args[])
{
//getting the size of the array
Scanner s = new Scanner(System.in);
int noe = s.nextInt();
int out[]=new int[noe];
int arr[] = new int[noe];
// getting the input array
for(int k=0;k<noe;k++)
{
arr[k]=s.nextInt();
}
int val1 = 1,val2=1;
for(int i=0;i<noe;i++)
{
int res=1;
for(int j=1;j<noe;j++)
{
if((i+j)>(noe-1))
{
int diff = (i+j)-(noe);
if(arr[diff]!=0)
{
res = res * arr[diff];
}
}
else
{
if(arr[i+j]!=0)
{
res= res*arr[i+j];
}
}
out[i]=res;
}
}
//printing result
System.out.print("Array of Product: [");
for(int l=0;l<out.length;l++)
{
if(l!=out.length-1)
{
System.out.print(out[l]+",");
}
else
{
System.out.print(out[l]);
}
}
System.out.print("]");
}
}
给你，简单干净的解决方案，复杂度为O(N):
int[] a = {1,2,3,4,5};
int[] r = new int[a.length];
int x = 1;
r[0] = 1;
for (int i=1;i<a.length;i++){
r[i]=r[i-1]*a[i-1];
}
for (int i=a.length-1;i>0;i--){
x=x*a[i];
r[i-1]=x*r[i-1];
}
for (int i=0;i<r.length;i++){
System.out.println(r[i]);
}
