Daily work entitled the twenty-seventh questions 20,200,329

/ **
*
* Question:
* Analyzing single chain acyclic intersect problem
* how to determine whether the two chain acyclic intersection, the first intersection intersection node returns a disjoint null is returned.
* If the two lists intersect acyclic, then starting from the intersection node, until the termination of this period of two lists, the two chains
* table sharing.
*
* Analysis:
* 1 1 list node start from scratch, go to the last node (not the end), the statistical list length 1 is denoted len1, the same.
* 1 records the last node in the list referred to end1 time.
*
* 2. list node 2 to start from scratch, go to the last node (not the end), the statistics list the length of the note 2 is len2, with
records list the last 2 nodes recorded as end2 * time.
*
.! * 3 if end1 = end2, illustrates two disjoint lists, can return null; if end1 = end2, say
* Description intersecting two lists, proceeds to step 4 to find the first intersection node.
*
* 4. If a long list, a list len1-len2 go first step, if the chain length of 2, 2 go first chain-LEN1 LEN2
* step. Then go together two lists, go with the process, the two lists first come first node that together, is the first
* a node intersect. (Because the shared node element after intersecting, so the extra length can not intersect a certain node)
*
* For example: a list of length 100, second length of the linked list 30, if it has been determined by Step 3 lists some phases 1 and 2 list
* Pay, then the next, linked lists 1 go first 70 steps, then go with the list 1 and 2 list, they will jointly enter the first
node * intersect.
*
* @Author snow pupil
*
* /

* Code

public class Node {
	public int value;
	public Node next;
	public Node(int data){
		this.value=data;
	}
}

　　

public class NodeLoop {

	Node currentHead1 = null;
	Node currentHead2 = null;
	int lengthNode1=0;
	int lengthNode2=0;
	int len = 0;
	public Node nodeLoop(Node head1,Node head2){
		
		if(head1==null&&head2==null){
			return null;
		}
		//遍历链表
		currentHead1 = head1;
		while(currentHead1!=null){
			lengthNode1++;
			currentHead1=currentHead1.next;
		}
		currentHead2 = head2;
		
		while(currentHead2!=null){
			lengthNode2++;
			currentHead2=currentHead2.next;
		}
		//相交判断
		if(currentHead1 !=currentHead2){
			return null;
		}else{
			if(lengthNode1>lengthNode2){
				int flag=0;
				len=lengthNode1-lengthNode2;
				currentHead1=head1;
				currentHead2=head2;
				
				while(currentHead1!=currentHead2){
					if(flag==len){
						currentHead1=currentHead1.next;
						currentHead2=currentHead2.next;
					}else{
						currentHead1=currentHead1.next;
					}
				}
				return currentHead2;
			}else if(lengthNode1<lengthNode2){
				int flag=0;
				len=lengthNode2-lengthNode1;
				currentHead1=head1;
				currentHead2=head2;
				
				while(currentHead1!=currentHead2){
					if(flag==len){
						currentHead1=currentHead1.next;
						currentHead2=currentHead2.next;
					}else{
						currentHead2=currentHead2.next;
					}
				}
				return currentHead1;
			}else{
				return head1;
			}
		}

	}	
}

　　

import java.util.Random;
import java.util.Scanner;


public class TestNodeLoop {

	public static void main(String[] args) {
		NodeLoop loop = new NodeLoop();
		TestNodeLoop test = new TestNodeLoop();
		
		Scanner sc = new Scanner(System.in);
		int length1;
		System.out.println("请输入链表长度:...");
		length1 = sc.nextInt();
		int length2;
		System.out.println("请输入链表长度:...");
		length2 = sc.nextInt();
		Node head1;
		Node head2;
		Node head3;
		head1=test.getNodeList(length1);
		head2=test.getNodeList(length2);
		test.showByTip(head1);
		test.showByTip(head2);
		int loopLength;
		System.out.println("请输入链表相交的节点位置:...");
		loopLength = sc.nextInt();
		head3=test.getLoopNode(head1,head2,loopLength);
		test.showByTip(head3);
		
		Node result = loop.nodeLoop(head2, head3);
		if(result!=null){
			System.out.println("链表相交相交元素如下:...");
			test.showByTip(result);
		}else{
			System.out.println("链表没有相交");
		}
		
		
		sc.close();
	}
	//获得相交链表
	private Node getLoopNode(Node head1, Node head2,int length) {
		int flag=0;
		Node current = null;
		current =head1;
		while(current!=null){
			flag++;
			if(flag==length){
				current.next=head2;
				return head1;
			}
			current=current.next;
		}
		System.out.println("链表不存在可以连接的位置，请增加元素...");
		return null;
	}

	//显示
	public void showByTip(Node head) {
		Node current = null;
		System.out.println("链表内的元素顺序显示如下：...");
		current=head;
		while(current!=null) {
			System.out.print(current.value+"\t");
			current=current.next;
		}
		System.out.println("");
	}
	//生成链表
	public Node getNodeList(int listLength) {
		Random rand = new Random();
		Node nodeList[]= new Node[listLength];
		for(int i=0;i<listLength;i++) {
			nodeList[i]= new Node(rand.nextInt(10)); 
		}
		for(int i=0;i<listLength-1;i++) {
			nodeList[i].next=nodeList[i+1];
		}
		return nodeList[0];
	}
}

　　

*运行