[Leetcode142]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。

这道题解法很巧妙，是个追击问题，首先是确定是否有环。然后在根据追击相遇点与表头的关系来寻找入环点。

python:

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def detectCycle(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        if head == None:
            return None
        if head.next == None:
            return None
        slow = head
        fast = head
        while(fast):
            slow = slow.next
            if fast.next:
                fast = fast.next.next
                if fast == slow:
                    break
            else:
                return None
        if fast == None :
            return None
        slow = head
        while(slow != fast):
            slow = slow.next
            fast = fast.next
        return slow

C++: 

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