15. Linked List Cycle && Linked List Cycle II

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;
    }
};