力扣142. 环形链表 II

Problem: 142. 环形链表 II

题目描述

思路

1.定义快慢指针fast和slow均指向链表的头节点，当fast没到达链表的结尾时，每次fast走两步，slow走一步；如果存在环则当slow和fast相遇时则退出循环；
2.若存在环，则再将slow指向头节点，再让slow和fast同时向后移动则再次相遇的位置就是环的起始点

复杂度

时间复杂度:

$O(n)$；其中$n$为链表的节点个数

空间复杂度:

$O(1)$

Code

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    /**
     * Find the starting position of the ring
     * 
     * @param head The head of linked list 
     * @return
     */
    ListNode *detectCycle(ListNode *head) {
        ListNode* fast = head;
        ListNode* slow = head;
        while (fast != nullptr && fast -> next != nullptr) {
            fast = fast -> next -> next;
            slow = slow -> next;
            if (fast == slow) {
                break;
            }
        }
        if (fast == nullptr || fast -> next == nullptr) {
            return nullptr;
        }

        slow = head;
        while (slow != fast) {
            fast = fast -> next;
            slow = slow -> next;
        }
        return slow;
    }
};