// ====================方法1====================
int LastRemaining_Solution1(unsigned int n, unsigned int m)
{
if(n < 1 || m < 1)
return -1;
unsigned int i = 0;
list<int> numbers;
for(i = 0; i < n; ++ i)
numbers.push_back(i);
list<int>::iterator current = numbers.begin();
while(numbers.size() > 1)
{
for(int i = 1; i < m; ++ i)
{
current ++;
if(current == numbers.end())
current = numbers.begin();
}
list<int>::iterator next = ++ current;
if(next == numbers.end())
next = numbers.begin();
-- current;
numbers.erase(current);
current = next;
}
return *(current);
}
// ====================方法2====================
int LastRemaining_Solution2(unsigned int n, unsigned int m)
{
if(n < 1 || m < 1)
return -1;
int last = 0;
for (int i = 2; i <= n; i ++)
last = (last + m) % i;
return last;
}