LeetCode 题解（144）: Linked List Cycle II

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题目：

Given a linked list, return the node where the cycle begins. If there is no cycle, returnnull.

Follow up:
Can you solve it without using extra space?

题解：

双指针fast进二，slow进一。当fast == slow时有环。

C++版：

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

Java版：

/** * Definition for singly-linked list. * class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { *         val = x; *         next = null; *     } * } */public class Solution {    public ListNode detectCycle(ListNode head) {        if(head == null || head.next == null)            return null;                    ListNode fast = head, slow = head;        while(fast != null && fast.next != null) {            fast = fast.next.next;            slow = slow.next;            if(fast == slow)                break;        }                if(fast == slow) {            ListNode p = head;            while(p != fast) {                p = p.next;                fast = fast.next;            }            return p;        }                return null;    }}

Python版：

# Definition for singly-linked list.# class ListNode:#     def __init__(self, x):#         self.val = x#         self.next = Noneclass Solution:    # @param head, a ListNode    # @return a list node    def detectCycle(self, head):        if head == None or head.next == None:            return None                    fast, slow = head, head        while fast != None and fast.next != None:            fast = fast.next.next            slow = slow.next            if fast == slow:                break                    if fast == slow:            p = head            while p != fast:                p = p.next                fast = fast.next            return p                    return None

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