160.intersection-of-two-linked-lists.cpp

/*
 * @lc app=leetcode id=160 lang=cpp
 *
 * [160] Intersection of Two Linked Lists
 *
 * https://leetcode.com/problems/intersection-of-two-linked-lists/description/
 *
 * algorithms
 * Easy (42.49%)
 * Likes:    4985
 * Dislikes: 564
 * Total Accepted:    596.9K
 * Total Submissions: 1.4M
 * Testcase Example:  '8\n[4,1,8,4,5]\n[5,6,1,8,4,5]\n2\n3'
 *
 * 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 = [5,6,1,8,4,5], 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 [5,6,1,8,4,5]. 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 = [1,9,1,2,4], 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
 * [1,9,1,2,4]. 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.
 * Each value on each linked list is in the range [1, 10^9].
 * Your code should preferably run in O(n) time and use only O(1) memory.
 * 
 * 
 */

// @lc code=start
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *getIntersectionNode1(ListNode *headA, ListNode *headB) {
        unordered_set<ListNode*> addSet;
        while (headA) {
            addSet.insert(headA);
            headA = headA->next;
        }
        
        while (headB) {
            if (addSet.count(headB))
                return headB;
            headB = headB->next;
        }
        return nullptr;
    }
    
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        ListNode* pA = headA;
        ListNode* pB = headB;
        
        while (pA != pB) {
            pA = pA == nullptr ? headB : pA->next;
            pB = pB == nullptr ? headA : pB->next;
        }
        
        return pA;
    }
};
// @lc code=end