两个单链表相交的一系列问题

请实现一个函数,如果两个链表相交,请返回相交的第一个节点;如果不想交,返回null即可。
要求:如果链表1的长度为N,链表2的长度为M,时间复杂度请达到O(N+M),额外空间复杂度请达到O(1)。

package chj;

public class Problem_11_FindFirstIntersectNode {


	public static class Node {
		public int value;
		public Node next;


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


	public static Node getIntersectNode(Node head1, Node head2) {
		if (head1 == null || head2 == null) {
			return null;
		}
		Node loop1 = getLoopNode(head1);
		Node loop2 = getLoopNode(head2);
		if (loop1 == null && loop2 == null) {
			return noLoop(head1, head2);
		}
		if (loop1 != null && loop2 != null) {
			return bothLoop(head1, loop1, head2, loop2);
		}
		return null;
	}


	public static Node getLoopNode(Node head) {
		if (head == null || head.next == null || head.next.next == null) {
			return null;
		}
		Node n1 = head.next; // n1 -> slow
		Node n2 = head.next.next; // n2 -> fast
		while (n1 != n2) {
			if (n2.next == null || n2.next.next == null) {
				return null;
			}
			n2 = n2.next.next;
			n1 = n1.next;
		}
		n2 = head; // n2 -> walk again from head
		while (n1 != n2) {
			n1 = n1.next;
			n2 = n2.next;
		}
		return n1;  //再次相交时(n1==n2),第一个入环的节点返回
	}


	public static Node noLoop(Node head1, Node head2) {
		if (head1 == null || head2 == null) {
			return null;
		}
		Node cur1 = head1;
		Node cur2 = head2;
		int n = 0;
		while (cur1.next != null) {
			n++;
			cur1 = cur1.next;
			//System.out.println("n===="+n);
			//System.out.println("cur1.value="+cur1.value);
		}
		System.out.println("链表1遍历完时,cur1指针对应的value值:"+cur1.value);
		while (cur2.next != null) {
			n--;
			cur2 = cur2.next;
			//System.out.println("n===="+n);
			//System.out.println("cur2.value="+cur2.value);
		}
		System.out.println("链表2遍历完时,cur2指针对应的value值:"+cur2.value);
		
		if (cur1 != cur2) {
			return null;
		}
		cur1 = n > 0 ? head1 : head2;
		cur2 = cur1 == head1 ? head2 : head1;
		//System.out.println("n的值 = "+n);
		n = Math.abs(n);
		while (n != 0) {
			n--;
			cur1 = cur1.next;
		}
		while (cur1 != cur2) {
			cur1 = cur1.next;
			cur2 = cur2.next;
		}
		return cur1;
	}


	public static Node bothLoop(Node head1, Node loop1, Node head2, Node loop2) {
		Node cur1 = null;
		Node cur2 = null;
		if (loop1 == loop2) {
			cur1 = head1;
			cur2 = head2;
			int n = 0;
			while (cur1 != loop1) {
				n++;
				cur1 = cur1.next;
			}
			System.out.println("链表1遍历完 时,cur1指针对应的value值:"+cur1.value);
			while (cur2 != loop2) {
				n--;
				cur2 = cur2.next;
			}
			cur1 = n > 0 ? head1 : head2;
			cur2 = cur1 == head1 ? head2 : head1;
			System.out.println("n = "+n);  // n = len_head1-len_head2 = 7-9 = -2
			n = Math.abs(n);
			while (n != 0) {
				n--;
				cur1 = cur1.next;
			}
			while (cur1 != cur2) {
				cur1 = cur1.next;
				cur2 = cur2.next;
			}
			return cur1;
		} else {
			cur1 = loop1.next;
			while (cur1 != loop1) {
				if (cur1 == loop2) {
					return loop1;
				}
				cur1 = cur1.next;
			}
			return null;
		}
	}


	public static void main(String[] args) {
		// 1->2->3->4->5->6->8->12->null
		Node head1 = new Node(1);
		head1.next = new Node(2);
		head1.next.next = new Node(3);
		head1.next.next.next = new Node(4);
		head1.next.next.next.next = new Node(5);
		head1.next.next.next.next.next = new Node(6);
		head1.next.next.next.next.next.next = new Node(8);
		head1.next.next.next.next.next.next.next = new Node(12);


		// 0->9->8->6->7->null
		Node head2 = new Node(0);
		head2.next = new Node(9);
		head2.next.next = new Node(8);
		head2.next.next.next = head1.next.next.next.next.next; // 8->6
		System.out.println("Common IntersectNode: " + getIntersectNode(head1, head2).value);
		//System.out.println(head2.next.next.next.next.value);
		System.out.println("==========分割线==========");

//		// 1->2->3->4->5->6->7->4...  head1为有环链表,节点为4
		head1 = new Node(1);  //Node head1 =new Node(1)
		head1.next = new Node(2);
		head1.next.next = new Node(3);
		head1.next.next.next = new Node(4);
		head1.next.next.next.next = new Node(5);
		head1.next.next.next.next.next = new Node(6);
		head1.next.next.next.next.next.next = new Node(7);
		head1.next.next.next.next.next.next = head1.next.next.next; // 7->4
//
//
//		// 0->9->8->2...   head2为有环节点,节点为4(其链表组成为:0->9->8->2->3->4->5->6->7->4... )
	    head2 = new Node(0);   //Node head1 =new Node(1)
		head2.next = new Node(9);
		head2.next.next = new Node(8);
		head2.next.next.next = head1.next; // 8->2
		System.out.println("head2.next.next.next.next.next的value值:"+head2.next.next.next.next.next.value);
		System.out.println("--IntersectNode--: "+ getIntersectNode(head1, head2).value);


		// 0->9->8->6->4->5->6.. ?      head2为有环节点
		head2 = new Node(0);
		head2.next = new Node(9);
		head2.next.next = new Node(8);
		head2.next.next.next = head1.next.next.next.next.next; // 8->6
		System.out.println("==IntersectNode==: "+ getIntersectNode(head1, head2).value);
		System.out.println("head2.next.next.next.next.next的value值为:"+head2.next.next.next.next.next.value);

	}


}

输出:

链表1遍历完时,cur1指针对应的value值:12
链表2遍历完时,cur2指针对应的value值:12
Common IntersectNode: 6
==========分割线==========
head2.next.next.next.next.next的value值:4
链表1遍历完 时,cur1指针对应的value值:4
n = -2
--IntersectNode--: 2
==IntersectNode==: 4
head2.next.next.next.next.next的value值为:5

posted on 2018-08-03 11:48  星辰之衍  阅读(117)  评论(0编辑  收藏

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