Ch2-5: find the beginning of loop in a circular linked list---two solutions

Given a circular linked list, implement an algorithm which returns node at the beginning of the loop.

DEFINITION

Circular linked list: A (corrupt) linked list in which a node’s next pointer points to an earlier node, so as to make a loop in the linked list.

EXAMPLE

Input: A -> B -> C -> D -> E -> C [the same C as earlier]

Output: C

这题有2中做法，书上的那个还没仔细看（2个指针，一个+2，一个+1，如果重复了，就说明有环。但是怎么找到环的头节点呢？关键是：

在重复的时候，把slow重新指向head,同时把fast的速度改为一，那么他们再次相遇的时候就是环路的头了）公式请参考hawstein。

I thinks the 2nd idea is simple as similar to early solution: use a hashmap to have key(node addr), if the addr is duplicate, it means there is a loop.

#include
#include
using namespace std;

typedef struct node{
int data;
node *next;
}node;

// initialize a linkedlist with a loop,
// the n is the total nodes, m is the starting
// node of the loop, still a singly linked list
node* init(int *a, int n, int m){
node *head, *p, *q;
for(int i=0; idata = a[i];
if(i==m) q = nd;
if(i==0){
head = p = nd;
continue;
}
p->next = nd;
p = nd;
}
p->next = q;
return head;
}

// seems like a math contest idea
node* loopstart0(node *head){   // see the explaination
node *fast = head, *slow=head;
while(fast || fast->next){
fast = fast->next->next;
slow = slow->next;
if(fast==slow)
break;
}
if(!fast || !fast->next) return NULL;
slow = head;
while(fast!=slow){
fast = fast->next;
slow = slow->next;
}
return fast;
}

// method 2: simple and easy to understand
map hashmap;

node* loopstart1(node *head){
while(head){
if(hashmap[head]) return head;
else{
hashmap[head] = true;
head = head->next;
}
}
return head;
}
int main(){
int n =6, m =2 ;// mdata<

输出结果：
Executing the program....
$demo
3