Given a linked list, determine if it has a cycle in it.
Follow up:
Can you solve it without using extra space?
Analysis:
Typical two pointer questions. Give two pointers fast and slow, fast run 2 steps and slow run 1 step. If there’s a cycle, finally slow == fast. If no cycle, fast == null || fast.next == null.
Time complexity: O(n)
Space complexity: O(1)
Code is below:
/**
* Definition for singly-linked list.
* class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
public boolean hasCycle(ListNode head) {
// O(n)
if (head == null || head.next == null) {
return false;
}
ListNode slow = head;
ListNode fast = head.next;
// while (slow != fast) {
// if (fast == null || fast.next == null) {
// return false;
// }
// slow = slow.next;
// fast = fast.next.next;
// }
// return true;
while (fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next.next;
if (slow == fast) {
return true;
}
}
return false;
}
}
Followup:
Given a linked list, return the node where the cycle begins. If there is no cycle, return null.Note: Do not modify the linked list.
Follow up:
Can you solve it without using extra space?
Analysis:
Follow the figure below:
链表头是X，环的第一个节点是Y，slow和fast第一次的交点是Z。各段的长度分别是a,b,c，如图所示。环的长度是L。slow和fast的速度分别是qs,qf。
第一次相遇时slow走过的距离：a+b，fast走过的距离：a+b+c+b。
因为fast的速度是slow的两倍，所以fast走的距离是slow的两倍，有 2(a+b) = a+b+c+b，可以得到a=c（这个结论很重要！）。
我们发现L=b+c=a+b，也就是说，从一开始到二者第一次相遇，循环的次数就等于环的长度。
我们已经得到了结论a=c，那么让两个指针分别从X和Z开始走，每次走一步，那么正好会在Y相遇！也就是环的第一个节点。
Time complexity: O(n)Space complexity: O(1)
Code is below:
/**
* Definition for singly-linked list.
* class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
public ListNode detectCycle(ListNode head) {
// O(n)
if (head == null || head.next == null) {
return null;
}
ListNode fast = head;
ListNode slow = head;
while (fast != null && fast.next != null) {
fast = fast.next.next;
slow = slow.next;
if (fast == slow) {
break;
}
}
if (fast != slow) {
return null;
}
fast = head;
while (fast != slow) {
fast = fast.next;
slow = slow.next;
}
return fast;
}
}
CodeBySteven 