代码题（15）— 环形链表

1、141. 环形链表

给定一个链表，判断链表中是否有环。

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

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

2、142. 环形链表 II

给定一个链表，返回链表开始入环的第一个节点。 如果链表无环，则返回 null。

说明：不允许修改给定的链表。

/**
 * 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(head == nullptr || head->next == nullptr)
            return nullptr;
        ListNode* slow = head;
        ListNode* fast = head->next->next;
        while(slow != fast)
        {
            if(fast != nullptr && fast->next != nullptr)
            {
                slow = slow->next;
                fast = fast->next->next;
            }
            else
                return nullptr;
        }
        fast = fast->next;
        int num = 1;
        while(slow != fast)
        {
            num++;
            fast = fast->next;
            
        }
        fast = head;
        for(int i=0;i<num;++i)
            fast = fast->next;
        slow = head;
        while(fast != slow)
        {
            slow = slow->next;
            fast = fast->next;
        }
        return slow;
    }
};