可以用两个指针来解决这个问题。先定义两个指针 P1 和 P2 指向链表的头结点。如果链表中环有 n 个结点,指针 P1 在链表上向前移动 n 步,然后两个指针以相同的速度向前移动。当第二个指针指向环的入口结点时,第一个指针已经围绕着环走了一圈又回到了入口结点。
剩下的问题就是如何得到环中结点的数目。我们在面试题 15 的第二个相关题目时用到了一快一慢的两个指针。如果两个指针相遇,表明链表中存在环。两个指针相遇的结点一定是在环中。可以从这个结点出发,一边继续向前移动一边计数,当再次回到这个结点时就可以得到环中结点数了。
private static class ListNode { private int val; private ListNode next; public ListNode() { } public ListNode(int val) { this.val = val; } @Override public String toString() { return val +""; } }
public class Test56 { private static class ListNode { private int val; private ListNode next; public ListNode() { } public ListNode(int val) { this.val = val; } @Override public String toString() { return val +""; } } public static ListNode meetingNode(ListNode head) { ListNode fast = head; ListNode slow = head; while (fast != null && fast.next != null) { fast = fast.next.next; slow = slow.next; if (fast == slow) { break; } } // 链表中没有环 if (fast == null || fast.next == null) { return null; } // fast重新指向第一个结点 fast = head; while (fast != slow) { fast = fast.next; slow = slow.next; } return fast; } public static void main(String[] args) { test01(); test02(); test03(); } // 1->2->3->4->5->6 private static void test01() { ListNode n1 = new ListNode(1); ListNode n2 = new ListNode(2); ListNode n3 = new ListNode(3); ListNode n4 = new ListNode(4); ListNode n5 = new ListNode(5); ListNode n6 = new ListNode(6); n1.next = n2; n2.next = n3; n3.next = n4; n4.next = n5; n5.next = n6; System.out.println(meetingNode(n1)); } // 1->2->3->4->5->6 // ^ | // | | // +--------+ private static void test02() { ListNode n1 = new ListNode(1); ListNode n2 = new ListNode(2); ListNode n3 = new ListNode(3); ListNode n4 = new ListNode(4); ListNode n5 = new ListNode(5); ListNode n6 = new ListNode(6); n1.next = n2; n2.next = n3; n3.next = n4; n4.next = n5; n5.next = n6; n6.next = n3; System.out.println(meetingNode(n1)); } // 1->2->3->4->5->6 <-+ // | | // +---+ private static void test03() { ListNode n1 = new ListNode(1); ListNode n2 = new ListNode(2); ListNode n3 = new ListNode(3); ListNode n4 = new ListNode(4); ListNode n5 = new ListNode(5); ListNode n6 = new ListNode(6); n1.next = n2; n2.next = n3; n3.next = n4; n4.next = n5; n5.next = n6; n6.next = n6; System.out.println(meetingNode(n1)); } }