Linked List Cycle II

Description

Given a linked list, return the node where the cycle begins.

If there is no cycle, return null.

Example
Given -21->10->4->5, tail connects to node index 1，return 10

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

Link

Lintcode_ladder

Method

1. x
2. x

Example

1. 1

/**
 * Definition of ListNode
 * class ListNode {
 * public:
 *     int val;
 *     ListNode *next;
 *     ListNode(int val) {
 *         this->val = val;
 *         this->next = NULL;
 *     }
 * }
 */
class Solution {
public:
    /**
     * @param head: The first node of linked list.
     * @return: The node where the cycle begins. 
     *           if there is no cycle, return null
     */
    ListNode *detectCycle(ListNode *head) {
        // write your code here
        // test whether it has loop
        if (!head) {
            return nullptr;
        }
        ListNode* slow = head;
        ListNode* fast = head->next;
        while (fast && fast->next) {
            if (slow == fast) {
                break;
            }
            slow = slow->next;
            fast = fast->next->next;
        }
        if (slow != fast) {
            return nullptr;
        }
        // find cycle point
        while (head != slow->next) {
            head = head->next;
            slow = slow->next;
        }
        return head;
    }
};

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