LeetCode 题解（144）: Linked List Cycle II

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 = None

class 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