算法基础之打印相交链表的公共节点

不考虑带环的情况...
这里面其实包含了两个问题：判断链表相交和找两个相交链表的第一个公共点。
思路：首先需要理解的是若两个链表相交，则 一定是呈Y型。所以我们分别遍历两个链表，找到最后一个节点，若它们相同，则一定相交。然后我们从后往前考虑，假设两个链表有n个公共节点，两个链表的长分别为l和m(设l>m)，那么我们先让较长的链表遍历l-m个。两个链表剩余部分长度是相等的(m)，则同步遍历比较，若相同，则输出。

import com.linkedlist.LinkedListReverse.Node;

/**
 * 打印相交链表的公共节点
 * 
 * @author aaron-han
 * 
 */
public class LinkedListCommonNodes {

	public static void main(String[] args) {
		Node a = new Node("NodeA");
		Node b = new Node("NodeB");
		Node c = new Node("NodeC");
		Node d = new Node("NodeD");
		Node e = new Node("NodeE");
		Node f = new Node("NodeF");
		Node g = new Node("NodeG");
		Node h = new Node("NodeH");

		// a->f->g->h
		a.next = f;
		f.next = g;
		g.next = h;

		// b->c->d->e->f->g->h
		b.next = c;
		c.next = d;
		d.next = e;
		e.next = f;
		f.next = g;
		g.next = h;

		System.out.println(isIntersected(a, c));
		printCommonNode(a, c);

	}

	// 判断是否相交
	public static boolean isIntersected(Node a, Node b) {
		Node headA = a;
		Node headB = b;
		while (headA.next != null) {
			headA = headA.next;
		}
		while (headB.next != null) {
			headB = headB.next;
		}
		if (headA == headB) {
			return true;
		}
		return false;
	}

	// 打印公共节点
	public static void printCommonNode(Node a, Node b) {
		int lengthA = getListLength(a);
		int lengthB = getListLength(b);

		Node longerList = a;
		Node shorterList = b;
		int lengthDiff = lengthA - lengthB;

		if (lengthDiff < 0) {
			longerList = b;
			shorterList = a;
			lengthDiff = lengthB - lengthA;
		}
		// 使较长的链表先遍历lengthDiff个
		for (int i = 0; i < lengthDiff; i++) {
			longerList = longerList.next;
		}
		// 此时等长，同步遍历
		while (longerList != null && shorterList != null) {
			if (longerList == shorterList) {
				System.out.println(longerList.name);
			}
			longerList = longerList.next;
			shorterList = shorterList.next;
		}
	}

	public static int getListLength(Node node) {
		int length = 0;
		if (node == null) {
			return length;
		}
		while (node != null) {
			node = node.next;
			length++;
		}
		return length;
	}

}