160. Intersection of Two Linked Lists.md

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:

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.

Solution

要比较的不是node.val而是node的内存，需要是同一个node才算重合，不然依然是不同的node，具体可见example1的1那个node。

自己写的code1，又慢又耗内存，好吧。

discussion里面的code2神仙解法，快并且不耗内存。

我们不需要知道两个链表的长度差，我们只需要保证两个指针可以同时指向交错的那个node👈核心

Code1

/**
 * 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) {
        HashSet<ListNode> set = new HashSet();
        ListNode node1 = headA;
        ListNode node2 = headB;
        
        while(node1!=null || node2!=null)
        {
            if(node1!=null)
            {
                if(set.contains(node1))
                    return node1;
                else
                    set.add(node1);
                node1 = node1.next;
            }
            if(node2!=null)
            {
                if(set.contains(node2))
                    return node2;
                else
                    set.add(node2);
                node2 = node2.next;
            }
        }
        return null;
    }
}

Code2

public class Solution {
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        // HashSet<ListNode> set = new HashSet();
        ListNode node1 = headA;
        ListNode node2 = headB;
        
        while(node1!=node2)
        {
            node1 = (node1==null)?headB:node1.next;
            node2 = (node2==null)?headA:node2.next;
        }
        return node1;
    }
}