Function:是否是环链表,采用快慢指针,一个走的快些一个走的慢些 如果最终相遇了就说明是环 就相当于在一个环形跑道里跑步,速度不一样的最终一定会相遇。 @author crossoverJie Date: 04/01/2018 11:33 @since JDK 1.8
| 9 | * @since JDK 1.8 |
| 10 | */ |
| 11 | public class LinkLoop { |
| 12 | |
| 13 | public static class Node{ |
| 14 | private Object data ; |
| 15 | public Node next ; |
| 16 | |
| 17 | public Node(Object data, Node next) { |
| 18 | this.data = data; |
| 19 | this.next = next; |
| 20 | } |
| 21 | |
| 22 | public Node(Object data) { |
| 23 | this.data = data ; |
| 24 | } |
| 25 | } |
| 26 | |
| 27 | /** |
| 28 | * 判断链表是否有环 |
| 29 | * @param node |
| 30 | * @return |
| 31 | */ |
| 32 | public boolean isLoop(Node node){ |
| 33 | Node slow = node ; |
| 34 | Node fast = node.next ; |
| 35 | |
| 36 | while (slow.next != null){ |
| 37 | Object dataSlow = slow.data; |
| 38 | Object dataFast = fast.data; |
| 39 | |
| 40 | //说明有环 |
| 41 | if (dataFast == dataSlow){ |
| 42 | return true ; |
| 43 | } |
| 44 | |
| 45 | //一共只有两个节点,但却不是环形链表的情况,判断NPE |
| 46 | if (fast.next == null){ |
| 47 | return false ; |
| 48 | } |
| 49 | //slow走慢点 fast走快点 |
| 50 | slow = slow.next ; |
| 51 | fast = fast.next.next ; |
| 52 | |
| 53 | //如果走的快的发现为空 说明不存在环 |
| 54 | if (fast == null){ |
| 55 | return false ; |
| 56 | } |
| 57 | } |
| 58 | return false ; |
| 59 | } |
| 60 | } |
nothing calls this directly
no outgoing calls
no test coverage detected