Created
December 24, 2013 10:39
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Linked List Cycle
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/** | |
* Given a linked list, determine if it has a cycle in it. | |
*/ | |
/** | |
* Definition for singly-linked list. | |
* class ListNode { | |
* int val; | |
* ListNode next; | |
* ListNode(int x) { | |
* val = x; | |
* next = null; | |
* } | |
* } | |
*/ | |
public class Solution{ | |
public boolean hasLoop(ListNode head){ | |
ListNode slow = head, fast = head; | |
while(slow != null && fast != null && fast.next != null){ | |
slow = slow.next; | |
fast = fast.next.next; | |
if(slow == fast) | |
return true; | |
} | |
return false; | |
} | |
} |
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/** | |
* Given a linked list, return the node where the cycle begins. If there is no cycle, return null. | |
*/ | |
public class Solution{ | |
public ListNode detectCycle(ListNode head) { | |
// The slow pointer advances one node at a time, while the fast pointer | |
// traverses twice as fast. If the list has loop in it, eventually the | |
// fast and slow pointer will meet at the same node. On the other hand, | |
// if the loop has no loop, the fast pointer will reach the end of list | |
// before the slow pointer does. | |
if(head == null || head.next == null || head.next.next == null) | |
return null; | |
ListNode slow = head.next; | |
ListNode fast = head.next.next; | |
while(slow != fast){ | |
if(fast.next == null || fast.next.next == null) | |
return null; | |
slow = slow.next; | |
fast = fast.next.next; | |
} | |
// If the program runs here, which means there is a loop, now advance | |
// one pointer to the head, and make them move in the same speed. | |
// When they meet again, it has to be the loop start point. | |
slow = head; // or fast = head | |
while(slow != fast){ | |
slow = slow.next; | |
fast = fast.next; | |
} | |
return slow; | |
} | |
} |
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