Linked List Cycle I 问有没有环
Given a linked list, determine if it has a cycle in it.
Challenge
Follow up:
Can you solve it without using extra space?
Example
Given -21->10->4->5, tail connects to node index 1, return true
public boolean hasCycle(ListNode head) {
if (head == null) {
return false;
}
ListNode slow = head;
ListNode fast = head.next;
while (fast != null && fast.next != null) {
if (slow == fast) {
return true;
}
slow = slow.next;
fast = fast.next.next;
}
return false;
}
Linked List Cycle II 问环的入口
Given a linked list, return the node where the cycle begins. If there is no cycle, return null.
Example
Given -21->10->4->5, tail connects to node index 1，返回10
Challenge
Follow up:
Can you solve it without using extra space?
Solution: 当在找环的过程中slow和fast相遇后，第一题直接return true就行了，这题需要找到环的入口。从slow.next到环的入口的距离等于从head到环的入口的距离。
public ListNode detectCycle(ListNode head) {
if (head == null) {
return null;
}
ListNode slow = head;
ListNode fast = head.next;
while (fast != null && fast.next != null && slow != fast) {
slow = slow.next;
fast = fast.next.next;
}
if (fast == null || fast.next == null) {
return null;
}
ListNode runner = head;
while (runner != slow.next) {
runner = runner.next;
slow = slow.next;
}
return runner;
}
