Linked List Cycle

Linked List Cycle - Problem

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 there is some node in the list that can be reached again by continuously following the next pointer. Internally, pos is used to denote the index of the node that 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.

Input & Output

Example 1 — Cycle Exists

$ Input: head = [3,2,0,-4], pos = 1

› Output: true

💡 Note: There is a cycle in the linked list, where the tail connects to the 1st node (0-indexed). The cycle is: 2 → 0 → -4 → back to 2.

Example 2 — Small Cycle

$ Input: head = [1,2], pos = 0

› Output: true

💡 Note: There is a cycle where the tail connects back to the 0th node. The cycle is: 1 → 2 → back to 1.

Example 3 — No Cycle

$ Input: head = [1], pos = -1

› Output: false

💡 Note: There is only one node and no cycle since pos = -1 indicates no connection back.

Constraints

The number of nodes in the list is in the range [0, 10⁴]
-10⁵ ≤ Node.val ≤ 10⁵
pos is -1 or a valid index in the linked-list

Visualization

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Asked in

A Amazon 42 M Microsoft 38 🍎 Apple 35 G Google 31

The key insight is that in a cycle, a fast-moving pointer will eventually catch a slow-moving pointer. The optimal approach uses Floyd's Cycle Detection Algorithm with two pointers moving at different speeds. Time: O(n), Space: O(1).

Common Approaches

✓ Dfs

⏱️ Time: N/A Space: N/A

Hash Set (Visited Nodes)

⏱️ Time: O(n) Space: O(n)

Traverse the linked list and store each visited node in a hash set. If we encounter a node that's already in the set, there's a cycle. If we reach the end (null), there's no cycle.

Two Pointers (Floyd's Algorithm)

⏱️ Time: O(n) Space: O(1)

Use two pointers moving at different speeds. The slow pointer moves one step at a time, while the fast pointer moves two steps. If there's a cycle, the fast pointer will eventually meet the slow pointer inside the cycle.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>

struct ListNode {
    int val;
    struct ListNode* next;
};

bool solution(struct ListNode* head) {
    struct ListNode* visited[10000];
    int visitedCount = 0;
    struct ListNode* current = head;
    
    while (current != NULL) {
        // Check if current node is already visited
        for (int i = 0; i < visitedCount; i++) {
            if (visited[i] == current) {
                return true;
            }
        }
        
        // Add current node to visited
        visited[visitedCount++] = current;
        current = current->next;
    }
    
    return false;
}

struct ListNode* arrayToLinkedList(int* arr, int size, int pos) {
    if (size == 0) {
        return NULL;
    }
    
    struct ListNode** nodes = (struct ListNode**)malloc(size * sizeof(struct ListNode*));
    
    // Create all nodes
    for (int i = 0; i < size; i++) {
        nodes[i] = (struct ListNode*)malloc(sizeof(struct ListNode));
        nodes[i]->val = arr[i];
        nodes[i]->next = NULL;
    }
    
    // Link nodes
    for (int i = 0; i < size - 1; i++) {
        nodes[i]->next = nodes[i + 1];
    }
    
    // Create cycle if pos is valid
    if (pos != -1 && pos < size) {
        nodes[size - 1]->next = nodes[pos];
    }
    
    struct ListNode* head = nodes[0];
    free(nodes);
    return head;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Parse input: [[arr], pos] or [arr, pos]
    int arr[1000];
    int size = 0;
    int pos = -1;
    
    // Simple parsing for the format [[3,2,0,-4],1]
    char* p = line;
    while (*p && *p != '[') p++; // Find first [
    if (*p == '[') p++;
    
    // Check if we have nested array
    if (*p == '[') {
        p++; // Skip second [
        // Parse array elements
        while (*p && *p != ']') {
            while (*p == ' ' || *p == ',') p++;
            if (*p == ']') break;
            if (*p == '-' || (*p >= '0' && *p <= '9')) {
                arr[size++] = (int)strtol(p, &p, 10);
            } else {
                p++;
            }
        }
        if (*p == ']') p++; // Skip first ]
        
        // Parse pos
        while (*p && (*p == ',' || *p == ' ')) p++;
        if (*p == '-' || (*p >= '0' && *p <= '9')) {
            pos = (int)strtol(p, &p, 10);
        }
    } else {
        // Single array format
        while (*p && *p != ']') {
            while (*p == ' ' || *p == ',') p++;
            if (*p == ']') break;
            if (*p == '-' || (*p >= '0' && *p <= '9')) {
                arr[size++] = (int)strtol(p, &p, 10);
            } else {
                p++;
            }
        }
    }
    
    struct ListNode* head = arrayToLinkedList(arr, size, pos);
    bool result = solution(head);
    
    if (result) {
        printf("true\n");
    } else {
        printf("false\n");
    }
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

✓ Linear Space

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