Circular linked list
Ring chain watch-Li buckle link
Subject content: 给定一个链表，判断链表中是否有环
Input and output example: true
1. Code implementation
Method 1: Use the non-repeatability of HashSet
public static boolean hasCycle2(ListNode head) {
//思路二：；利用HashSet的无重复性
if(head == null )return false;
Set<ListNode> set = new HashSet<>();
while (head != null){
if (set.contains(head)){
return true;
}
set.add(head);
head = head.next;
}
return false;
}
Method two: fast and slow pointer
public static boolean hasCycle2(ListNode head) {
//思路一：比较好想到的思路就是快慢指针
if(head == null )return false;
ListNode quick = head.next;
ListNode cur = head;
while (cur != quick){
if (quick == null || quick.next ==null){
return false;
}
cur = cur.next;
quick = quick.next.next;
}
return true;
}
2. Advanced circular linked list (1)
- If it is a circular linked list, how to know the length of the loop ;
Analysis of the train of thought: Use the above speed pointer to continue moving forward in one encounter, the number of moves in the middle is the length;
//求环形链表的长度(扩展)
public static int hasCycleLength(ListNode head) {
if(head == null )return 0;
ListNode quick = head.next;
ListNode cur = head;
while (cur != quick){
if (quick == null || quick.next ==null){
return 0;
}
cur = cur.next;
quick = quick.next.next;
}
//循环中出来说明第一次相遇；那么继续
int count = 1;//因为有环，最少也有一个节点
quick = quick.next;
while (cur != quick){
cur = cur.next;
quick = quick.next.next;
count++;
}
return count;
}
3. Advanced circular linked list (2)
- Return the starting node of the circular linked list (that is, the inbound node) ;
Problem-solving ideas:重新拿一个节点从头开始遍历，另外一个节点p1从相遇点前进，则相遇点就是入环节点
public static ListNode firstNode(ListNode head){
//双指针的方法，一个的移动速度是另外一个移动速度的二倍
ListNode p1 = head;
ListNode p2 = head;
ListNode p3 = head;
System.out.println(p3);
while(p2 != null && p2.next != null) {
//这里只需要判断p2就可以，因为p2移动的快
p1 = p1.next;
p2 = p2.next.next;
if (p1 == p2){
break;//相等则有环，类似于小学的追击问题，就像操场跑步一样，一直跑下去，跑的快一定会追上跑的慢的
}
}
//出来要判断一下
if (p2 != null){
//如果p2不等于空，则说明是相遇了，才从循环中出来的
//则在次进行循环
while (true){
p1 = p1.next;
p3 = p3.next;
if (p1 == p3){
System.out.println(p3);
return p3;
}
}
}
return null;
}