LC 417. Linked List Cycle II

题目描述

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.

参考答案

/**
 * 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 == NULL || head->next == NULL) return NULL;
        ListNode* fp = head;
        ListNode* sp = head;
        bool isCycle = false;

        while(fp != NULL && sp != NULL){
            fp = fp -> next;
            if(sp -> next == NULL) return NULL;
            sp = sp->next->next;
            if(fp == sp) {
                isCycle = true;
                break;
            }
        }

        if(!isCycle) return NULL;

        fp = head;
        while(fp != sp) {
            fp = fp->next;
            sp = sp->next;
        }
        return fp;

    }
};

答案注释

想象 fast 的速度是一步，slow 的速度是不动。那么，常规情况下，他俩在a点集合，在a点再次相遇。

但由于slow 摸鱼了，晚来了，fast已经走到b点了，slow才和fast集合，一起出发。a-b 就是所求的 循环开始点 到 列表开始点的距离。

更详细解释，可参考：https://www.cnblogs.com/hiddenfox/p/3408931.html