题目

输入两个链表，找出它们的第一个公共节点。

public static class ListNode{public int val;public ListNode next;public ListNode(int val) {this.val = val;}
}

分析

首先遍历两链表的长度。在第二次遍历的时候，在较长的链表上先走若干步，步数为两链表之差的绝对值，接着同时在两个链表遍历，找到的第一个相同的节点就是它们的第一个公共节点。

放码

public class FindTheSameNode {public ListNode find(ListNode head1, ListNode head2) {int length1 = count(head1);int length2 = count(head2);int diff = Math.abs(length1 - length2);ListNode pLong = length1 > length2 ? head1 : head2, pShort = length1 > length2 ? head2 : head1;while(diff > 0) {pLong = pLong.next;diff--;}while(pLong != null && pShort != null && pLong != pShort) {pLong = pLong.next;pShort = pShort.next;}return  pLong == pShort ? pLong : null;}public int count(ListNode head) {int result = 0;ListNode p = head;while(p != null) {result++;p = p.next;}return result;}}

测试

import static org.junit.Assert.*;
import org.junit.Test;
import com.lun.util.SinglyLinkedList.ListNode;public class FindTheSameNodeTest {private static ListNode list1;private static ListNode list2;private static ListNode same;static {same = new ListNode(6);same.next = new ListNode(7);list1 = new ListNode(1);list1.next = new ListNode(2);list1.next.next = new ListNode(3);list1.next.next.next = same;list2 = new ListNode(4);list2.next = new ListNode(5);list2.next.next = same;}@Testpublic void testFind() {		FindTheSameNode fn = new FindTheSameNode();assertEquals(6, fn.find(list1, list2).val);assertEquals(same, fn.find(list1, list2));}@Testpublic void testCount() {FindTheSameNode fn = new FindTheSameNode();assertEquals(5, fn.count(list1));assertEquals(4, fn.count(list2));}}