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:

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.

解题要点：

遍历两根链表，直到他们有相同的节点，返回它。

/**
 * 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) {
        ListNode l1 = headA, l2 = headB;
        while(l1 != l2){
            l1 = (l1 == null) ? headB : l1.next;
            l2 = (l2 == null) ? headA : l2.next;
        }
        return l1;
    }
}