題目

給定一個連結串列，如果連結串列中存在環，則返回到連結串列中環的起始節點的值，如果沒有環，返回null。

給出 -21->10->4->5, tail connects to node index 1，返回10

分析

Python程式碼

"""
Definition of ListNode
class ListNode(object):

    def __init__(self, val, next=None):
        self.val = val
        self.next = next
"""
class Solution:
    """
    @param head: The first node of the linked list.
    @return: The node where the cycle begins. 
                if there is no cycle, return null
    """
    def detectCycle(self, head):
        # write your code here
        if head == None or head.next == None:
            return None
            
        slow = head
        fast = head
        
        while fast != None and fast.next != None:
            slow = slow.next
            fast = fast.next.next
            if slow == fast:
                break
            
        if fast != None and fast == slow:
            fast = head
            while fast != slow:
                slow = slow.next
                fast = fast.next
            return fast
        
        return None
 

C++程式碼

/**
103 帶環連結串列 II

給定一個連結串列，如果連結串列中存在環，則返回到連結串列中環的起始節點的值，如果沒有環，返回null。

您在真實的面試中是否遇到過這個題？ Yes
樣例
給出 -21->10->4->5, tail connects to node index 1，返回10

*/

/**
 * Definition of ListNode
 * class ListNode {
 * public:
 *     int val;
 *     ListNode *next;
 *     ListNode(int val) {
 *         this->val = val;
 *         this->next = NULL;
 *     }
 * }
 */
class Solution {
public:
    /**
     * @param head: The first node of linked list.
     * @return: The node where the cycle begins. 
     *           if there is no cycle, return null
     */
    ListNode *detectCycle(ListNode *head) {
        // write your code here
        if(head == NULL || head->next ==NULL)
        {
            return NULL;
        }//if
        
        ListNode *slow = head, *fast = head;
        while(fast && fast->next)
        {
            slow = slow->next;
            fast = fast->next->next;
            if(slow == fast)
            {
                break;
            }//if
        }//while
        
        if(fast && fast == slow)
        {
            fast = head;
            while(fast != slow)
            {
                fast = fast->next;
                slow = slow->next;
            }//while
            
            return fast;
        }//if
        return NULL;
        
    }
};