142. Linked List Cycle II

problem description

Example 1:

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

1

Example 2:

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.

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

2

Example 3:

Input: head = [1], pos = -1
Output: tail connected nothing
Explanation: There is no cycle in the linked list.

3

algorithm thought

这题和上一题很像，用一样的算法就能解决，利用数学证明可以得知，当两个快慢指针相交时，他们到连接点的长度就是起始点到连接点的长度。这时候，一个指针从起始点出发，一个从相交点出发，两个指针相交之后就能知道连接点是哪个

code

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

algorithm analysis

和上一题一样，时间复杂度O(n),空间复杂度O(1)