Solution 7: 判断两链表是否相交

问题描述

RT.

 

解决思路

(1) 两链表都是单向链表：判断两链表的末尾节点是否相同；

(2) 两链表中一个有环，一个没环：不可能相交；

(3) 两链表都有环：slow-fast双指针方法。

 

程序

 

public class ListIntersection {

	// two single list

	public boolean isIntersectionOfTwoSingleList(ListNode l1, ListNode l2) {

		if (l1 == null || l2 == null) {

			return false;

		}



		// whether the end of two list is same

		ListNode endOfList1 = getEndOfList(l1);

		ListNode endOfList2 = getEndOfList(l2);



		return endOfList1 == endOfList2;

	}



	private ListNode getEndOfList(ListNode head) {

		if (head == null) {

			return null;

		}



		ListNode node = head;

		while (node.next != null) {

			node = node.next;

		}

		return node;

	}



	// two list with cycle

	public boolean isIntersectionOfTwoListWithCycle(ListNode l1, ListNode l2) {

		if (l1 == null || l2 == null) {

			return false;

		}



		ListNode slow = l1, fast = l2;



		while (fast.next != null || fast != null || slow != null) {

			slow = slow.next;

			fast = fast.next.next;

			if (slow == fast) {

				return true;

			}

		}



		return false;

	}

}

 

Follow up

求出两个链表相交的第一个节点（如果存在的话）。

 

(1) 两条单链表

a. 求出两条链表的长度及长度之差diff，然后设立两个指针指向两链表的头结点，其中指向长链表头结点的指针向前移动diff步；

b. 然后同时移动两指针，直到所指节点相同（地址相同）为止，否则返回null。

 

(2) 两条链表有环

首先slow-fast，直到相遇为止，其中任意一个指针指回其头结点，然后slow和fast指针同时移动，直到相遇，相遇的节点为第一个相交的节点。

（注意：可能有两个相交的节点）

 

程序

public class ListIntersection2 {

	// two single list

	public ListNode getFirstIntersectionNodeOfSingleList(ListNode l1,

			ListNode l2) {

		ListNode longNode = l1, shortNode = l2;



		int len1 = getLenOfList(l1);

		int len2 = getLenOfList(l2);



		if (len1 < len2) {

			longNode = l2;

			shortNode = l1;

		}



		int diff = Math.abs(len1 - len2);



		// long move diff steps

		while (diff > 0) {

			longNode = longNode.next;

			--diff;

		}



		while (longNode != null && shortNode != null) {

			if (longNode == shortNode) {

				return longNode;

			}

			longNode = longNode.next;

			shortNode = shortNode.next;

		}



		return null;

	}



	private int getLenOfList(ListNode head) {

		ListNode node = head;

		int len = 0;



		while (node != null) {

			++len;

			node = node.next;

		}



		return len;

	}



	// two list with cycle

	public ListNode getFirstIntersectionNodeOfCycleList(ListNode l1, ListNode l2) {

		if (l1 == null || l2 == null) {

			return null;

		}



		ListNode slow = l1, fast = l2;



		while (fast.next != null || fast != null || slow != null) {

			if (fast == slow) {

				break;

			}

			slow = slow.next;

			fast = fast.next.next;

		}



		if (fast == null || slow == null) {

			return null;

		}



		slow = l1;

		while (slow != fast) {

			slow = slow.next;

			fast = fast.next;

		}

		

		return slow;

	}

}