P14-环形链表-快慢指针
2021-05-13 本文已影响0人
YonchanLew
//环形链表
/*
* 给定一个链表,判断链表中是否有环
* 如果链表中有某个节点,可以通过连续跟踪next指针再次到达该节点,则链表中存在环
* 如果链表中存在环,返回true,否则false
* */
public class P14 {
static class ListNode{
int val;
ListNode next;
public ListNode(int val, ListNode next){
this.val = val;
this.next = next;
}
}
public static void main(String[] args) {
ListNode node5 = new ListNode(5, null);
ListNode node4 = new ListNode(4, node5);
ListNode node3 = new ListNode(3, node4);
ListNode node2 = new ListNode(2, node3);
ListNode node1 = new ListNode(1, node2);
node5.next = node3;
System.out.println(hasCycle(node1));
System.out.println(hasCycle2(node1));
}
private static boolean hasCycle(ListNode head) {
//空间 O(n)
Set<ListNode> set = new HashSet<>();
//时间 O(n)
while(head != null){
//添加失败会返回false
if(!set.add(head)){
return true;
}
head = head.next;
}
return false;
}
//双指针,快慢指针,龟兔赛跑
private static boolean hasCycle2(ListNode head) {
if(head == null || head.next == null){
return false;
}
ListNode slow = head;
ListNode quick = head.next;
//时间 O(n)
while(slow != quick){
if(quick == null || quick.next == null){
return false;
}
slow = slow.next;
//走得快,同时减少遍历次数
//快指针提前进入环,迟早和慢指针相遇
quick = quick.next.next;
}
return true;
}
}