Check if a singly linked list is cyclic

For a single table as follows, how do we determine whether it is a cycle or not.

We let the two pointers, the fast and slow pointers respectively, point to the head node, let the fast pointer move two nodes at a time, and the slow pointer move only one node at a time.

Since fast travels faster and slow travels slower, there must be a meeting node in the ring, then the length of the path traveled by slow is len + x, and the path traveled by fast is nR + len + x , n is the number of turns for fast, greater than or equal to 1; R is the length of the ring.
Since fast is twice as fast as slow, the path taken by fast is twice as long as slow.
Thus nR + len + x = 2len + 2x; thus, len = nR - x;
at this time, we will let the two pointers x and y start from the head node and the junction respectively, and the first encounter point will arrive The entry node of the ring.

Then the idea of ​​​​this algorithm is introduced here. We deduce len=nR-x, so that they must meet at the entry point.

//C实现代码
//判断是否成环，如果成环，返回环入口结点
Node * isCylization(Node * node){
    if(node == NULL) return NULL;
    Node * fast = node;
    Node * slow = node;
    while(1){
        fast = fast->next?fast->next:NULL;
        if(!fast)return NULL;
        fast = fast->next?fast->next:NULL;
        if(!fast)return NULL;
        slow = slow->next?slow->next:NULL;
        if(!slow)return NULL;
        if(slow == fast) break;
    }
    Node * x = node;
    Node * y = slow;
    for(;x == y;x=x->next;y=y->next);
    return x;
}
int main(){

    return 0;
}