给定一个单链表，如何找到带有数字的模

Question

Given a decimal number represented as list from least significant bit to most significant bit how should one find mod with k if the list were a number.

Example :- number - 1234567

list representation of number: 7->6->5->4->3->2->1

k = 7

ans = 1234567%7 = 5

In python and java I could have converted the list to a number BigInteger(java) and have taken the mod. But I wanted to do it without using BigInteger.

I have tried calculating the mod while iterating the list

total=0
for digit in numberList:
     total = (total * 10 + digit) % k
return total

But this is not going to work.

Answer 1

You can calculate the remainder of the i'th digit multiplied by 10^i for each digit. Then sum all these remainders and find the remainder of the sum :

int mul = 1;
int remainder = 0;
for (int digit : digitArray)
{
    remainder += (digit * mul) % k;
    mul *= 10;
}
remainder = remainder % k;

For 1234567%7 , this loop basically computes :

(7 % 7 + 60 % 7 + 500 % 7 + 4000 % 7 + 30000 % 7 + 200000 % 7 + 1000000 % 7) % 7 =
(0 + 4 + 3 + 3 + 5 + 3 + 1) % 7 = 19 % 7 = 5

Answer 2

Your code is close, but you need to process your number list in reverse. I'll illustrate with a simple Python list .

num = 1234567

# Construct numberList
numberList = []
n = num
while n:
    n, d = divmod(n, 10)
    numberList.append(d)

k = 7
print(num, numberList, num % k)

# Find numberList mod k
r = 0
for d in reversed(numberList):
    r = (10 * r + d) % k

print(r)

output

1234567 [7, 6, 5, 4, 3, 2, 1] 5
5

Answer 3

Think of the basic property of modular arithmetic :
(a*b)%m=(a%m * b%m ) % m
(a+b)%m=(a%m + b%m ) % m
Now, to find modulo of 1234567
1234567 % m = (1234 * 1000 + 567)%m =((1234 * 1000)%m + 567%m) %m
Now, take a little small no. to better understand this trick
12345 =( ( (1 * 10 + 2) * 10 + 3) * 10 + 4) *10 +5
Now,
12345 % m = (((((((1*10)%m + 2)*10)%m + 3)*10)%m + 4)*10 + 5)%m

Now to find modulo of a number that is represented by a linked list its actually easy
list1: 1->2->3->4->5.......
take variable result and initialize with 0
ie, result=0 and just traverse the list and do:
result = (result*10 )% m + node->data
see the code below it is too simple

int result=0;
while(list1){
    int x=list1->data;
    result=(result*10)%m + x;
    list1=list1->next;
}
result=result%m;
return result;