Leetcode(160) Intersection of Two Linked Lists

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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:

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begin to intersect at node c1.

Example 1:

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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:

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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:

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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.

解法

用一个HashSet存储节点即可，遍历完两个链表，如果有相交节点，那么第一个冲突的点就是交点啦

具体代码如下：

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

        }
        return null;
    }
}