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.
Method:

Find the sizes of the arrays first. then move the pointer of the longer array until the sizes are the same. Then scan both arrays at the same time until they meet.

Solution:

Time: O(n)
Space: O(1)

/**
* class ListNode {
*     public int val;
*     public ListNode next;
*     ListNode(int x) { val = x; next = null; }
* }
*/
public class Solution {
public ListNode getIntersectionNode(ListNode a, ListNode b) {
int aSize = 0;
ListNode aTmp = a;
while (aTmp != null) {
aSize ++;
aTmp = aTmp.next;
}
int bSize = 0;
ListNode bTmp = b;
while (bTmp != null) {
bSize ++;
bTmp = bTmp.next;
}
if (aSize > bSize) {
int diff = aSize - bSize;
while (diff > 0) {
a = a.next;
diff --;
}
} else {
int diff = bSize - aSize;
while (diff > 0) {
b = b.next;
diff --;
}
}
while (a != b) {
a = a.next;
b = b.next;
}
return a;
}
}