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?

分析：和Linked List Cycle类似，还是用map。

用时：60ms

/**
 * 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) {
        map<ListNode*, bool> m;
        while(head){
            if(m.find(head) == m.end()) m[head] = true;
            else return head;
            head = head->next;
        }
        return NULL;
    }
};

 

同时，对于Linked List Cycle中的较优方法，同样适用于本题。当fast和slow指针相遇时，令设指针slow2 = head，那么slow2和slow一定会在相遇的地方重合。

用时：16ms

/**
 * 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) {
        ListNode *fast = head, *slow = head;
        while(fast && fast->next){
            fast = fast->next->next;
            slow = slow->next;
            if(fast == slow){
                ListNode* slow2 = head;
                while(slow){
                    if(slow2 == slow) return slow;
                    slow = slow->next;
                    slow2 = slow2->next;
                    
                }
            }
        }
        return NULL;
    }
};