[LeetCode] 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.

Thought process:

Iterate both linked lists twice:

1. Get the lengths of both lists.
2. Increment the longer list by the difference between the lists lengths. Iterate both lists again until their pointers collide.

Solution 1 (Length):

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47

/** * 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) { int lengthA = 0; int lengthB = 0; ListNode currentA = headA; ListNode currentB = headB; while (currentA != null) { currentA = currentA.next; lengthA++; } while (currentB != null) { currentB = currentB.next; lengthB++; } currentA = headA; currentB = headB; if (lengthA > lengthB) { for (int i = 0; i < lengthA - lengthB; i++) { currentA = currentA.next; } } else { for (int i = 0; i < lengthB - lengthA; i++) { currentB = currentB.next; } } while (currentA != currentB) { currentA = currentA.next; currentB = currentB.next; } return currentA; } }

The question can also be thought as a pursue problem. Iterate both lists together, when one pointer finishes its list, switch to the head of the other list, and continue iterating until the end. If the lists intersect, the pointers will meet at the start of the intersection. If not, both pointers will iterate length a + b and break out of the loop.

Solution 2 (Pursue):

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37

/** * 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 != null || p2 != null) { if (p1 == null) { p1 = headB; } if (p2 == null) { p2 = headA; } if (p1 == p2) { return p1; } p1 = p1.next; p2 = p2.next; } return null; } }

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