0160_intersection_of_two_linked_lists.cc

/**
 * Author: Lanqing Ye
 * Date: 2019-10-02
 */
/* 

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.

*/ 

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */

// Fun One: 
class Solution {
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        if(headA == NULL || headB == NULL)  return nullptr;
        
        while(headA != NULL){
            ListNode *tmp = headB;
            while(tmp != NULL){
                if(headA == tmp){
                    return headA;
                }
                tmp = tmp -> next;
            }
            
            headA = headA -> next;
        }
        return nullptr;
    }
}; 

// Fun Two:
class Solution {
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        ListNode *NodeA = headA;
        ListNode *NodeB = headB;
        while(NodeA != NodeB){
            if(NodeA != NULL)
                NodeA = NodeA -> next;
            else
                NodeA = headB;
            if(NodeB != NULL)
                NodeB = NodeB -> next;
            else
                NodeB = headA;
        }
        return NodeA;
    }
};