Reputation: 35
I am trying to decrement a number, which can be infinitely long and is represented in a vector, by 1. As a small example:
vector<int> v1 = {5, 0, 0, 0};
After subtracting one from the end, the result should be:
vector<int> v1 = {4, 9, 9, 9};
This is my current code:
int size = v1.size();
bool carry = false;
for (int i = size - 1; i > 0; i--) {
if (v1.at(i) == 0) {
v1.at(i) = 9;
if (v1.at(0) == 1) {
v1.at(0) = 0;
}
carry = true;
} else {
v1.at(i) -= 1;
carry = false;
}
}
if (carry == true && v1.at(0) == 0) {
v1.erase(v1.begin());
} else if (carry == true) {
v1.at(0) -= 1;
}
return v1;
When I test it, everything works fine, except numbers like 11119. They turn out to be 00019. Is there anything I could tweak?
Upvotes: 0
Views: 58
Reputation: 9
A subtraction as you tried to do, could be done like this:
#include <iterator>
std::vector<int> v1 = {5, 0, 0, 0};
for (std::vector<int>::reverse_iterator i = v1.rbegin(); i != v1.rend(); i++) {
*i -= 1;
if (*i < 0) {
*i = 9;
}
else {
break;
}
}
Upvotes: 0
Reputation: 206627
It seems to me that you didn't think through the logic clearly.
Here's what needs to happen.
If the last number is 0, it needs to be changed to 9 and a carry has to be maintained.
Repeat until no carry needs to be maintained.
That logic is best implemented using a do - while
loop. Here's what I came up with.
int size = v.size();
bool carry = true;
int i = size - 1;
do
{
if (v.at(i) == 0)
{
v.at(i) = 9;
}
else
{
v.at(i)--;
carry = false;
}
--i;
}
while ( carry == true && i >= 0);
Here's a complete program
#include <iostream>
#include <vector>
void test(std::vector<int> v)
{
int size = v.size();
bool carry = true;
int i = size - 1;
do
{
if (v.at(i) == 0)
{
v.at(i) = 9;
}
else
{
v.at(i)--;
carry = false;
}
--i;
}
while ( carry == true && i >= 0);
for ( auto item : v )
{
std::cout << item << " ";
}
std::cout << std::endl;
}
int main()
{
test({1, 1, 1, 1, 9});
test({5, 0, 0, 0, 0});
}
and its output
1 1 1 1 8
4 9 9 9 9
See it working at https://ideone.com/lxs1vz.
Upvotes: 1