圆圈中最后剩下的数字

N个人围一圈报数，数到3 退出，最后一个是几号
（1）自定义链表
typedef struct Node
{
int value;
struct Node* next;

}Node ,*Link;
Link creat(int n){
Link node=new Node;
node->value=n-1;
Link pre,rear;
rear=pre=node;
for(int i=n-2;i>=0;i--){
node =new Node;
node->value=i;
node->next=pre;
pre=node;
}
rear->next=pre;
return pre;
}
int LastRemaining(int m,Link list){
Link p=list,q;
int i=0;
while(p->next!=p){
i++;
if(i==m-1){
q=p->next;
p->next=q->next;
delete q;
i=0;
}
p=p->next;
}
int last=p->value;
return last;
}
int _tmain(int argc, _TCHAR* argv[])
{
Link list=creat(5);
int result=LastRemaining(3,list);
printf("%d\n",result);
}
（2）使用std::list
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);
}
时间O（mn），空间O（n）
// ====================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;
}
时间O（n），空间O(1)；