# Intersection of Two Linked Lists

160. Intersection of Two Linked Lists

LintCode-380.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.

```
/**
* 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;
}
int lenA = getLen(headA);
int lenB = getLen(headB);
if (lenA > lenB) {
while (lenA > lenB) {
headA = headA.next;
lenA--;
}
} else {
while (lenA < lenB) {
headB = headB.next;
lenB--;
}
}
while (headA != null) {
if (headA == headB) {
return headA;
}
headA = headA.next;
headB = headB.next;
}
return null;
}
int getLen(ListNode node) {
int len = 0;
while (node != null) {
len++;
node = node.next;
}
return len;
}
}
```

```
public class Solution {
public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
if (headA == null || headB == null) {
return null;
}
ListNode a = headA, b = headB;
while (a != b) {
a = (a == null) ? headB : a.next;
b = (b == null) ? headA : b.next;
}
return a;
}
}
```

Hope this helps,

Michael