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
根据Billz的回答——抱歉我不能评论，但这里是一个正确处理列表中重复项的scala版本，可能是O(n):
val list1 = List(1, 7, 3, 3, 4, 4)
val view = list1.view.zipWithIndex map { x => list1.view.patch(x._2, Nil, 1).reduceLeft(_*_)}
view.force
返回:
List(1008, 144, 336, 336, 252, 252)
其他回答
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)
多基因润滑剂方法的一个解释是:
诀窍是构造数组(在4个元素的情况下):
{ 1, a[0], a[0]*a[1], a[0]*a[1]*a[2], }
{ a[1]*a[2]*a[3], a[2]*a[3], a[3], 1, }
这两种方法都可以在O(n)中分别从左右边开始。
然后，将两个数组逐个元素相乘，得到所需的结果。
我的代码看起来是这样的:
int a[N] // This is the input
int products_below[N];
int p = 1;
for (int i = 0; i < N; ++i) {
products_below[i] = p;
p *= a[i];
}
int products_above[N];
p = 1;
for (int i = N - 1; i >= 0; --i) {
products_above[i] = p;
p *= a[i];
}
int products[N]; // This is the result
for (int i = 0; i < N; ++i) {
products[i] = products_below[i] * products_above[i];
}
如果你也需要空间中的解是O(1)，你可以这样做(在我看来不太清楚):
int a[N] // This is the input
int products[N];
// Get the products below the current index
int p = 1;
for (int i = 0; i < N; ++i) {
products[i] = p;
p *= a[i];
}
// Get the products above the current index
p = 1;
for (int i = N - 1; i >= 0; --i) {
products[i] *= p;
p *= a[i];
}
我的第一次尝试，用Python。O (2 n):
def product(l):
product = 1
num_zeroes = 0
pos_zero = -1
# Multiply all and set positions
for i, x in enumerate(l):
if x != 0:
product *= x
l[i] = 1.0/x
else:
num_zeroes += 1
pos_zero = i
# Warning! Zeroes ahead!
if num_zeroes > 0:
l = [0] * len(l)
if num_zeroes == 1:
l[pos_zero] = product
else:
# Now set the definitive elements
for i in range(len(l)):
l[i] = int(l[i] * product)
return l
if __name__ == "__main__":
print("[0, 0, 4] = " + str(product([0, 0, 4])))
print("[3, 0, 4] = " + str(product([3, 0, 4])))
print("[1, 2, 3] = " + str(product([1, 2, 3])))
print("[2, 3, 4, 5, 6] = " + str(product([2, 3, 4, 5, 6])))
print("[2, 1, 2, 2, 3] = " + str(product([2, 1, 2, 2, 3])))
输出:
[0, 0, 4] = [0, 0, 0]
[3, 0, 4] = [0, 12, 0]
[1, 2, 3] = [6, 3, 2]
[2, 3, 4, 5, 6] = [360, 240, 180, 144, 120]
[2, 1, 2, 2, 3] = [12, 24, 12, 12, 8]
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;
}
技巧:
使用以下方法:
public int[] calc(int[] params) {
int[] left = new int[n-1]
in[] right = new int[n-1]
int fac1 = 1;
int fac2 = 1;
for( int i=0; i<n; i++ ) {
fac1 = fac1 * params[i];
fac2 = fac2 * params[n-i];
left[i] = fac1;
right[i] = fac2;
}
fac = 1;
int[] results = new int[n];
for( int i=0; i<n; i++ ) {
results[i] = left[i] * right[i];
}
是的，我确定我错过了一些I -1而不是I，但这是解决它的方法。
