141、142 Linked List Cycle & II

141、 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?

思路
非常简单的一道题，如果链表有环，则在遍历的时候永远不会结束，一直在环内转圈。
所以，设置快慢指针，如果相遇，则说明有环。

代码（C++）

/** * 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)     {        if (NULL == head || NULL == head->next)            return false;        ListNode* slow = head;        ListNode* fast = head;        while (fast->next && fast->next->next)        {            slow = slow->next;            fast = fast->next->next;            if (slow == fast)                return true;        }        return false;    }};

142、Linked List Cycle II
Given a linked list, return the node where the cycle begins. If there is no cycle, return null.

Note: Do not modify the linked list.

思路
在上一题的基础上，slow和fast相遇后，slow和head同时出发，每次一个步长，相遇的时候就是交点。

代码（C++）

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    ListNode *detectCycle(ListNode *head)     {        if (NULL == head || NULL == head->next)            return false;        ListNode* slow = head;        ListNode* fast = head;        while (fast->next && fast->next->next)        {            slow = slow->next;            fast = fast->next->next;            if (slow == fast)            {                ListNode* tmp = head;                while (tmp != slow)                {                    tmp = tmp->next;                    slow = slow->next;                }                return tmp;            }        }        return NULL;        }};