142. Linked List Cycle II

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每日一题 2019 - 05 - 08

题目：

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:

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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.

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Example 2:

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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.

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Example 3:

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

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

解法：

这个题让找出链表存在的环的入口，思路十分的巧妙，也即通过快慢双指针法进行求解，快指针每次运动两步，慢指针每次运行一步，当快慢指针相遇时候，此时慢指针到达入口的距离一定等于链表起始点到达入口的距离（这里是有严格的数学推导的)，此时再从链表头部设置一个指针开始追赶慢指针，一定就能找到入口所在；

代码：

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/**
 * 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 *fast = head ;
        ListNode *low = head ;
        ListNode *first = head ;
        if(head == NULL)
        {
            return NULL ;
        }
        ListNode *temp = head ;
        int i = 0 ;
        while(temp)
        {
            temp = temp -> next ;
            i ++ ;
            if(i > 2)
            {
                break ;
            }
        }
        if(i == 1)
        {
            return NULL ;
        }
        while(fast -> next != NULL && fast -> next -> next != NULL)
        {
            fast = fast -> next -> next ;
            low = low -> next ;
            if(fast == low)
            {
                while(first != low)
                {
                    first = first -> next ;
                    low = low -> next ;
                }
                return first ;
            }
        }
        return NULL ;
    }
};
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