Monday, May 1, 2017

160. Intersection of Two Linked Lists

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

For example, the following two linked lists: 
A:          a1 → a2
                   ↘
                     c1 → c2 → c3
                   ↗            
B:     b1 → b2 → b3
begin to intersect at node c1.

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.



Solution:

If two linked lists have a intersection, it should look like this:










The length of A is x + z, and the length of B is y + z.

Let's traverse step by step from A and B.

We find if B travels y + z to the end of the list, and travels x more steps from the beginning of linked list A, it travels y + z + x steps.

If A travels x + z to the end of the list, and travels y more steps from the beginning of linked list B, it also travels y + z + x steps.

And both of them will arrive the intersection.



Code:


/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        if (headA == null || headB == null) {
            return null;
        }
        
        ListNode p1 = headA;
        ListNode p2 = headB;
        
        while (p1 != p2) {
            p1 = p1 == null ? headB : p1.next;
            p2 = p2 == null ? headA : p2.next;
        }
        
        return p1;
    }
}