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
这是O(n²)但f#太漂亮了
List.fold (fun seed i -> List.mapi (fun j x -> if i=j+1 then x else x*i) seed)
[1;1;1;1;1]
[1..5]
其他回答
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]);
}
下面是一个C实现 O(n)时间复杂度。 输入
#include<stdio.h>
int main()
{
int x;
printf("Enter The Size of Array : ");
scanf("%d",&x);
int array[x-1],i ;
printf("Enter The Value of Array : \n");
for( i = 0 ; i <= x-1 ; i++)
{
printf("Array[%d] = ",i);
scanf("%d",&array[i]);
}
int left[x-1] , right[x-1];
left[0] = 1 ;
right[x-1] = 1 ;
for( i = 1 ; i <= x-1 ; i++)
{
left[i] = left[i-1] * array[i-1];
}
printf("\nThis is Multiplication of array[i-1] and left[i-1]\n");
for( i = 0 ; i <= x-1 ; i++)
{
printf("Array[%d] = %d , Left[%d] = %d\n",i,array[i],i,left[i]);
}
for( i = x-2 ; i >= 0 ; i--)
{
right[i] = right[i+1] * array[i+1];
}
printf("\nThis is Multiplication of array[i+1] and right[i+1]\n");
for( i = 0 ; i <= x-1 ; i++)
{
printf("Array[%d] = %d , Right[%d] = %d\n",i,array[i],i,right[i]);
}
printf("\nThis is Multiplication of Right[i] * Left[i]\n");
for( i = 0 ; i <= x-1 ; i++)
{
printf("Right[%d] * left[%d] = %d * %d = %d\n",i,i,right[i],left[i],right[i]*left[i]);
}
return 0 ;
}
输出
Enter The Size of Array : 5
Enter The Value of Array :
Array[0] = 1
Array[1] = 2
Array[2] = 3
Array[3] = 4
Array[4] = 5
This is Multiplication of array[i-1] and left[i-1]
Array[0] = 1 , Left[0] = 1
Array[1] = 2 , Left[1] = 1
Array[2] = 3 , Left[2] = 2
Array[3] = 4 , Left[3] = 6
Array[4] = 5 , Left[4] = 24
This is Multiplication of array[i+1] and right[i+1]
Array[0] = 1 , Right[0] = 120
Array[1] = 2 , Right[1] = 60
Array[2] = 3 , Right[2] = 20
Array[3] = 4 , Right[3] = 5
Array[4] = 5 , Right[4] = 1
This is Multiplication of Right[i] * Left[i]
Right[0] * left[0] = 120 * 1 = 120
Right[1] * left[1] = 60 * 1 = 60
Right[2] * left[2] = 20 * 2 = 40
Right[3] * left[3] = 5 * 6 = 30
Right[4] * left[4] = 1 * 24 = 24
Process returned 0 (0x0) execution time : 6.548 s
Press any key to continue.
下面是一个使用c#的函数式示例:
Func<long>[] backwards = new Func<long>[input.Length];
Func<long>[] forwards = new Func<long>[input.Length];
for (int i = 0; i < input.Length; ++i)
{
var localIndex = i;
backwards[i] = () => (localIndex > 0 ? backwards[localIndex - 1]() : 1) * input[localIndex];
forwards[i] = () => (localIndex < input.Length - 1 ? forwards[localIndex + 1]() : 1) * input[localIndex];
}
var output = new long[input.Length];
for (int i = 0; i < input.Length; ++i)
{
if (0 == i)
{
output[i] = forwards[i + 1]();
}
else if (input.Length - 1 == i)
{
output[i] = backwards[i - 1]();
}
else
{
output[i] = forwards[i + 1]() * backwards[i - 1]();
}
}
我不完全确定这是O(n)，因为所创建的Funcs是半递归的，但我的测试似乎表明它在时间上是O(n)。
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;
}
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]));
