LeetCode - Linked List Cycle II

Linked List Cycle II

2014.1.13 21:43

Given a linked list, return the node where the cycle begins. If there is no cycle, return null.

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

Solution:

　　This problem doesn't only ask you to check if there is a loop in the list, but also to find out where you enter the loop.

　　It is a bit tricky to solve this problem, if you are too accustomed to that "chasing method" that we mentioned in Linked List Cycle.

　　Since you only have a "next" pointer, what you can do is to check if $ptr or $ptr->next satisfy some specific property.

　　See the following linked list:

　　

　　Node 2 is the entrance of the loop. Both node 1 and node 4 points to it. If you start traversing the list, you are sure to reach node 2 first and node 4 later. If there is a $ptr pointing to node 4, you reach $ptr->next before you reach $ptr,right? In a normal linked list, this will never happen. That's how we use abnormal condition to detect this abnormal node.

　　This problem doesn't use a chasing strategy with two pointers. But it requires one-by-one probing for all nodes until the right one is found, with each probing having O(n) time complexity. Therefore, the overall time complexity is O(n^2), space complexity is O(1).

Accepted code:

// 1CE, 2TLE, 1AC, foolish mistake..
/**
 * 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) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        if(head == nullptr){
            return nullptr;
        }
        
        ListNode *ptr, *res;
        
        res = head;
        while(res != nullptr){
            ptr = head;
            // Forgot to check nullptr, added by the way later.
            while(ptr != res->next && ptr != nullptr){
                if(ptr == res){
                    // ptr reaches res first, not what we expect
                    break;
                }
                // 1TLE here, forgot to move forward...
                // 1CE here, ';' is missing!!!!
                ptr = ptr->next;
            }
            if(ptr == res->next){
                // $ptr reaches res->next first, that means $res->next is the start of the loop
                // while $res is the end of the loop, thus $ptr = $res->next is the result we want
                return ptr;
            }else{
                // 1TLE here, forgot to move forward..
                res = res->next;
            }
        }
        
        return nullptr;
    }
};