LeetCode_160. Intersection of Two Linked Lists

160. Intersection of Two Linked Lists

Easy

Write a program to find the node at which the intersection of two singly linked lists begins.

For example, the following two linked lists:

begin to intersect at node c1.

Example 1:

Input: intersectVal = 8, listA = [4,1,8,4,5], listB = [5,0,1,8,4,5], skipA = 2, skipB = 3
Output: Reference of the node with value = 8
Input Explanation: The intersected node's value is 8 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [4,1,8,4,5]. From the head of B, it reads as [5,0,1,8,4,5]. There are 2 nodes before the intersected node in A; There are 3 nodes before the intersected node in B.

Example 2:

Input: intersectVal = 2, listA = [0,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1
Output: Reference of the node with value = 2
Input Explanation: The intersected node's value is 2 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [0,9,1,2,4]. From the head of B, it reads as [3,2,4]. There are 3 nodes before the intersected node in A; There are 1 node before the intersected node in B.

Example 3:

Input: intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2
Output: null
Input Explanation: From the head of A, it reads as [2,6,4]. From the head of B, it reads as [1,5]. Since the two lists do not intersect, intersectVal must be 0, while skipA and skipB can be arbitrary values.
Explanation: The two lists do not intersect, so return null.

Notes:

• If the two linked lists have no intersection at all, return null.
• The linked lists must retain their original structure after the function returns.
• You may assume there are no cycles anywhere in the entire linked structure.
• Your code should preferably run in O(n) time and use only O(1) memory.

package leetcode.easy;

/**
 * Definition for singly-linked list. public class ListNode { int val; ListNode
 * next; ListNode(int x) { val = x; next = null; } }
 */
public class IntersectionOfTwoLinkedLists {
	public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
		if (null == headA || null == headB) {
			return null;
		} else {
			ListNode pA = headA;
			ListNode pB = headB;
			while (pA != pB) {
				if (pA != null) {
					pA = pA.next;
				} else {
					pA = headB;
				}
				if (pB != null) {
					pB = pB.next;
				} else {
					pB = headA;
				}
			}
			return pA;
		}
	}

	@org.junit.Test
	public void test0() {
		ListNode a1 = new ListNode(1);
		ListNode a2 = new ListNode(2);
		ListNode b1 = new ListNode(1);
		ListNode b2 = new ListNode(2);
		ListNode b3 = new ListNode(3);
		ListNode c1 = new ListNode(1);
		ListNode c2 = new ListNode(2);
		ListNode c3 = new ListNode(3);
		a1.next = a2;
		b1.next = b2;
		b2.next = b3;
		a2.next = c1;
		b3.next = c1;
		c1.next = c2;
		c2.next = c3;
		c3.next = null;
		System.out.println(getIntersectionNode(a1, b1).val);
	}

	@org.junit.Test
	public void test1() {
		ListNode a1 = new ListNode(4);
		ListNode a2 = new ListNode(1);
		ListNode b1 = new ListNode(5);
		ListNode b2 = new ListNode(0);
		ListNode b3 = new ListNode(1);
		ListNode c1 = new ListNode(8);
		ListNode c2 = new ListNode(4);
		ListNode c3 = new ListNode(5);
		a1.next = a2;
		b1.next = b2;
		b2.next = b3;
		a2.next = c1;
		b3.next = c1;
		c1.next = c2;
		c2.next = c3;
		c3.next = null;
		System.out.println(getIntersectionNode(a1, b1).val);
	}

	@org.junit.Test
	public void test2() {
		ListNode a1 = new ListNode(0);
		ListNode a2 = new ListNode(9);
		ListNode a3 = new ListNode(1);
		ListNode b1 = new ListNode(3);
		ListNode c1 = new ListNode(2);
		ListNode c2 = new ListNode(4);
		a1.next = a2;
		a2.next = a3;
		a3.next = c1;
		b1.next = c1;
		c1.next = c2;
		c2.next = null;
		System.out.println(getIntersectionNode(a1, b1).val);
	}

	@org.junit.Test
	public void test3() {
		ListNode a1 = new ListNode(2);
		ListNode a2 = new ListNode(6);
		ListNode a3 = new ListNode(4);
		ListNode b1 = new ListNode(1);
		ListNode b2 = new ListNode(5);
		a1.next = a2;
		a2.next = a3;
		a3.next = null;
		b1.next = b2;
		b2.next = null;
		System.out.println(getIntersectionNode(a1, b1));
	}
}