160. Intersection of Two Linked Lists

Description

Write a program to find the node at which the intersection of two singly linked lists begins.

For example, the following two linked lists:

begin to intersect at node c1.

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.

Idea

Two Points:

• Maintain two pointers pA and pB initialized at the head of A and B, respectively. Then let them both traverse through the lists, one node at a time.
• When pA reaches the end of a list, then redirect it to the head of B (yes, B, that’s right.); similarly when pB reaches the end of a list, redirect it the head of A.
• If at any point pA meets pB, then pA/pB is the intersection node.

Solution

/**
 * 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) {
        if (!headA || !headB) {
            return NULL;
        }
        int end = 0;
        ListNode *pa = headA;
        ListNode *pb = headB;
        while (pa->next && pb->next) {
            pa = pa->next;
            pb = pb->next;
        }
        ListNode *tmp, *temp;
        if (pa->next) {
            tmp = pa;
            temp = pb;
            pa = headA;
            pb = headB;
            while (tmp->next) {
                tmp = tmp->next;
                pa = pa->next;
            }
            if (tmp != temp) {
                return NULL;
            }
            while (pa != pb) {
                pa = pa->next;
                pb = pb->next;
            }
        } else {
            tmp = pb;
            temp = pa;
            pa = headA;
            pb = headB;
            while (tmp->next) {
                tmp = tmp->next;
                pb = pb->next;
            }
            if (tmp != temp) {
                return NULL;
            }
            while (pa != pb) {
                pa = pa->next;
                pb = pb->next;
            }
        }
        
        return pa;
    }
};