Question

02.07. Intersection of Two Linked Lists

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Description

Given two (singly) linked lists, determine if the two lists intersect. Return the inter­ secting node. Note that the intersection is defined based on reference, not value. That is, if the kth node of the first linked list is the exact same node (by reference) as the jth node of the second linked list, then they are intersecting.

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.

Solutions

Solution 1

# Definition for singly-linked list.
# class ListNode:
#   def __init__(self, x):
#     self.val = x
#     self.next = None


class Solution:
  def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode:
    cur1, cur2 = headA, headB
    while cur1 != cur2:
      cur1 = headB if cur1 is None else cur1.next
      cur2 = headA if cur2 is None else cur2.next
    return cur1

/**
 * 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 cur1 = headA, cur2 = headB;
    while (cur1 != cur2) {
      cur1 = cur1 == null ? headB : cur1.next;
      cur2 = cur2 == null ? headA : cur2.next;
    }
    return cur1;
  }
}

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *   int val;
 *   ListNode *next;
 *   ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
  ListNode* getIntersectionNode(ListNode* headA, ListNode* headB) {
    ListNode* cur1 = headA;
    ListNode* cur2 = headB;
    while (cur1 != cur2) {
      cur1 = cur1 ? cur1->next : headB;
      cur2 = cur2 ? cur2->next : headA;
    }
    return cur1;
  }
};

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *   Val int
 *   Next *ListNode
 * }
 */
func getIntersectionNode(headA, headB *ListNode) *ListNode {
  cur1, cur2 := headA, headB
  for cur1 != cur2 {
    if cur1 == nil {
      cur1 = headB
    } else {
      cur1 = cur1.Next
    }
    if cur2 == nil {
      cur2 = headA
    } else {
      cur2 = cur2.Next
    }
  }
  return cur1
}

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *   this.val = val;
 *   this.next = null;
 * }
 */

/**
 * @param {ListNode} headA
 * @param {ListNode} headB
 * @return {ListNode}
 */
var getIntersectionNode = function (headA, headB) {
  let cur1 = headA;
  let cur2 = headB;
  while (cur1 != cur2) {
    cur1 = cur1 ? cur1.next : headB;
    cur2 = cur2 ? cur2.next : headA;
  }
  return cur1;
};