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。
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)
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
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)