leetcode Linked List Cycle II

问题：Given a linked list, return the node where the cycle begins. If there is no cycle, return null

解决：快慢指针，判断是否有环，然后找进入环的节点



假设在Z点相遇，则快指针走了a+b+c+b,慢指针走了a+b,又因为快指针走了慢指针的两倍，所以2(a+b)=a+b+c+b,推出a=c,所以相遇之后让一个指针从z点出发，另外一个指针从X点出发，每次走一步，相遇的点就是Y点，代码如下
struct ListNode {
int val;
ListNode *next;
ListNode(int x) : val(x), next(NULL) {}
};

class Solution {
public:
ListNode *detectCycle(ListNode *head) {
ListNode *fastp = head, *slowp = head;
while(1){
if(fastp == NULL ||fastp->next == NULL) return NULL;
fastp = fastp->next->next;
slowp = slowp->next;
if(slowp == fastp) break;
}

slowp = head;
while(slowp != fastp){
slowp = slowp->next;
fastp = fastp->next;
}
return slowp;
}
};