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Leetcode Linked List Cycle problem solution

In this Leetcode Linked List Cycle problem solution we have Given head, the head of a linked list, determine if the linked list has a cycle in it. There is a cycle in a linked list if some node in the list can be reached again by continuously following the next pointer. Internally, pos is used to denote the index of the node that the tail's next pointer is connected to. Note that pos is not passed as a parameter. Return true if there is a cycle in the linked list. Otherwise, return false.

Leetcode Linked List Cycle problem solution

Problem solution in Python.

class Solution:
    def hasCycle(self, head: ListNode) -> bool:
        if head is None:
            return False
        while start is not None and end is not None and end.next is not None:
            if start==end:
                return True
        return False

Problem solution in Java.

public class Solution {
    public boolean Cycle(ListNode root) {
        if(root == null || root.next == null){
            return false;
        ListNode slow = root;
        ListNode fast = root;
        while(fast != null && fast.next != null){
            fast = fast.next.next;
            slow = slow.next;
            if( slow == fast){
                return true;
        return false;

Problem solution in C++.

class Solution
	bool hasCycle(ListNode* head)
		if(NULL == head) return false;
		ListNode* p = head;
		ListNode* q = head;
			p = p->next;
			q = q->next;
			if(NULL != q) q = q->next;
			if(NULL == q || NULL == p) return false;
			else if(p == q) return true;

Problem solution in C.

struct ListNode *inverse(struct ListNode *head) {
    struct ListNode *prev = NULL, *cur = head, *next = head; 
    while (next!=NULL) {
        cur = next;
        next = cur->next;
        cur->next = prev;
        prev = cur;
        if (next==head) {
            next->next = cur;
            return next;
    return cur;
bool hasCycle(struct ListNode *head) {
    if (head==NULL || head->next==NULL) {
        return 0;
    struct ListNode *tail = inverse(head);
    int r = (head==tail);
    return r; 

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