(链表 双指针) leetcode 142. Linked List Cycle II

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

To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle in the linked list.

Note: Do not modify the linked list.

 

Example 1:

Input: head = [3,2,0,-4], pos = 1
Output: tail connects to node index 1
Explanation: There is a cycle in the linked list, where tail connects to the second node.

Example 2:

Input: head = [1,2], pos = 0
Output: tail connects to node index 0
Explanation: There is a cycle in the linked list, where tail connects to the first node.

Example 3:

Input: head = [1], pos = -1
Output: no cycle
Explanation: There is no cycle in the linked list.

 

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

 

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这个题是不仅是判断链表中是否存在环，还要返回环的开始的位置。

用双指针，与141. Linked List Cycle 的类似，关于怎么返回这个位置，可以参考这个大佬的博客：http://www.cnblogs.com/hiddenfox/p/3408931.html

C++代码：

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