Intersection of Two Linked Lists

/* Write a program to find the node at which

* the intersection of two singly linked lists begins.

* This solution runs O(n) time.

*

* Basic idea:

* 1. Traverse the two linked list to find length M and N;

* 2. Get back to the heads, the skip |M - N| nodes on the longer list;

* 3. Now walk through both lists together until you find the common node.

*

* Important notes:

* When submitting to leetcode, the debug info

* System.out.println() should be commented,

* since it is very time consuming!

*/

public class Solution {

public ListNode getIntersectionNode(ListNode headA, ListNode headB) {

/* My answer starts here */

ListNode ptrA = headA;

ListNode ptrB = headB;

int lengthA = 0;

int lengthB = 0;

int i;

// get the length of list A

while (null != ptrA) {

//System.out.println("ptrA: " + ptrA.val);

lengthA++;

ptrA = ptrA.next;

}

ptrA = headA; // pointer back to the head

// get the length of list B

while (null != ptrB) {

//System.out.println("ptrB: " + ptrB.val);

lengthB++;

ptrB = ptrB.next;

}

ptrB = headB; // pointer back to the head

// skip the first |M - N| nodes of the longer list

if (lengthA >= lengthB) {

for (i = 0; i < lengthA - lengthB; i++) {

ptrA = ptrA.next;

}

} else {

for (i = 0; i < lengthB - lengthA; i++) {

ptrB = ptrB.next;

}

}

// compare one by one of the remaining nodes

while (null != ptrA) {

if (ptrA.val == ptrB.val) {

return ptrA; // find the intersection

}

ptrA = ptrA.next;

ptrB = ptrB.next;

}

return null; // no intersection

/* My answer ends here */

}

}