Linked List Cycle
Given a linked list, determine if it has a cycle in it.
Follow up: Can you solve it without using extra space?
说明：两个指针不同步长。
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
bool hasCycle(ListNode *head) {
ListNode *p1, *p2;
p1 = p2 = head;
while(p2 && p2->next && p2->next->next) {
p2 = p2->next->next;
p1 = p1->next;
if(p1 == p2) return true;
}
return false;
}
};
Linked List Cycle II
Given a linked list, return the node where the cycle begins. If there is no cycle, return null.
Follow up: Can you solve it without using extra space?
说明：在上题基础上，将一个指针放到链表头，步长都设为1，相遇节点。（可以计算）
class Solution {
public:
ListNode *detectCycle(ListNode *head) {
ListNode *p1, *p2;
p1 = p2 = head;
while(p2 && p2->next && p2->next->next) {
p2 = p2->next->next;
p1 = p1->next;
if(p1 == p2) {
p1 = head;
while(p1 != p2) {
p1 = p1->next;
p2 = p2->next;
}
return p1;
}
}
return NULL;
}
};