Intersection of Two Linked Lists

题目地址：https://leetcode.com/problems/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:

A:          a1 → a2                   ↘                     c1 → c2 → c3                   ↗            B:     b1 → b2 → b3

begin to intersect at node c1.

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.

找出两个链表相交的第一个节点，加入两个链表有相交的节点，那么从这个节点的以后两个链表都是相同的，不止是val相同，地址也是相同的，如果用两层for循环，肯定能找到这个节点，不过这么做的时间复杂度是O(m*n)，其中m与n分别是两个链表的长度。

有没有遍历一次就能找到相交节点呢，当然是可以的，我们可以把两个链表的每个节点依次散列在哈希表中，然后再去检查第一次出现冲突的节点，那么这个节点就是相交节点。

public class IntersectionOfTwoLinkedLists {    public static ListNode getIntersectionNode(ListNode headA, ListNode headB) {        if (headA == null || headB == null)            return null;        HashMap<ListNode, Integer> map = new HashMap<>();        ListNode p = headA;        while (p != null) {            map.put(p, 0);            p = p.next;        }        p = headB;        while (p != null) {            if (map.containsKey(p))                break;            p = p.next;        }        return p;    }    public static void main(String[] args) {        ListNode a1 = new ListNode(1);        a1.next = new ListNode(2);        a1.next.next = new ListNode(3);        a1.next.next.next = new ListNode(4);        a1.next.next.next.next = new ListNode(5);        ListNode b1 = new ListNode(1);        b1.next = new ListNode(2);        b1.next.next = a1.next.next;        System.out.println(getIntersectionNode(a1, b1).val);    }}