Aracade Intro #57 fileNaming. Algorithm,알고리즘,Codefights,CodeSignal,코드파이트,코드시그널,예제,문제해결능력,example,c++,java,재귀,recursive
Recursive solotion
재귀풀이
Q.
You are given an array of desired filenames in the order of their creation. Since two files cannot have equal names, the one which comes later will have an addition to its name in a form of (k), where k is the smallest positive integer such that the obtained name is not used yet.
Return an array of names that will be given to the files.
만든 순서에 따라서 만들어진 적당한 파일이름의 배열을 입력받았다. 두개의 파일은 같은 이름을 가질 수 없기 때문에, 나중에 오는 것들은 (k)가 붙는데, k 는 아직 쓰이지 않은 최소의 정수이다.
파일들에 주어질 이름들의 배열을 만들어라
e.g.
Input -> names = ["doc", "doc", "image", "doc(1)", "doc"]
Output -> fileNaming(names) = ["doc", "doc(1)", "image", "doc(1)(1)", "doc(2)"]
//Process
//1. Input fileNames vector
//2. Iterate from begin to the end
// 2.1.
//1. 파일 이름들 배열을 입력받는다.
//2. 시작부터 끝까지 반복한다.
// 2.1.
Code.. Lemme see code!!!!!
코드.. 코드를 보자!!!!!
std::vector<std::string> fileNaming(std::vector<std::string> names) {
std::vector<std::string> returnVector;
for (int i = 0; i < names.size(); ++i)
{
int count = 0;
makeName(&returnVector, names[i], count);
}
return returnVector;
}
void makeName(std::vector<std::string> *returnVector, string name, int count) {
if (hasSame(returnVector, name))
{
if (count == 0)
{
++count;
name = name + "(" + itos(count) + ")";
}
else
{
++count;
int digitCount = digitNumber(count-1);
name.replace(name.length()-1-digitCount, digitCount, itos(count));
}
makeName(returnVector, name, count);
}
else
{
returnVector->push_back(name);
}
}
int digitNumber(int number) {
int digitCount = 0;
while (number > 0)
{
++digitCount;
number /= 10;
}
return digitCount;
}
bool hasSame(vector<string> *returnVector, string name) {
bool answer = false;
for (int i = 0; i < returnVector->size(); ++i)
{
if (name == returnVector->at(i))
{
answer = true;
}
}
return answer;
}
string itos(int integer) {
std::stringstream ss;
ss << integer;
return ss.str();
}
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