给你两个单链表的头节点 headA 和 headB ，请你找出并返回两个单链表相交的起始节点。如果两个链表没有交点，返回 nu11。

以下有两种解决方法:

• 一种是用Map,利用其key值唯一的方法去判断(也可以使用set,set在add时,已存在的元素会返回false,不存在的返回true),但是此种方法会导致额外的空间消耗;
• 另外一种是利用双指针,获取两个链表中的长度,将最长的起始部位和最短的起始部分相等,一起遍历.

    static class ListNode{
        private int val;
        private ListNode node;

        public ListNode(int val, ListNode node) {
            this.val = val;
            this.node = node;
        }

        @Override
        public String toString() {
            return "ListNode{" +
                    "val=" + val +
                    ", node=" + node +
                    '}';
        }
    }

    public static void main(String[] args) {
        ListNode node5 = new ListNode(5, null);
        ListNode node4 = new ListNode(4, node5);
        ListNode node3 = new ListNode(3, node4);
        ListNode node2 = new ListNode(2, node3);
        ListNode node1 = new ListNode(1, node2);
        ListNode head3 = new ListNode(3, node3);
        ListNode head2 = new ListNode(2, head3);
        ListNode head1 = new ListNode(1, head2);
        System.out.println("相交链表元素为:" + getIntersectionNode(head1, node1));
        System.out.println("相交链表元素为:" + getIntersectionNode2(head1, node1));
    }

    //相交链表
    private static ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        if (headA == null || headB == null) {
            return null;
        }
        int a = 0, b = 0, c = 0;
        ListNode nodea = headA, nodeb = headB;
        while (nodea != null) {
            a++;
            nodea = nodea.node;
        }
        while (nodeb != null) {
            b++;
            nodeb = nodeb.node;
        }
        nodea = headA;
        nodeb = headB;
        if (a < b) {
            c = b - a;
            for (int i = 0; i < c; i++) {
                nodeb = nodeb.node;
            }
        } else {
            c = a - b;
            for (int i = 0; i < c; i++) {
                nodea = nodea.node;
            }
        }
        while (nodea != null && nodeb != null) {
            if (nodea == nodeb)
                return nodea;
            nodea = nodea.node;
            nodeb = nodeb.node;
        }
        return null;
    }

    private static ListNode getIntersectionNode2(ListNode headA, ListNode headB) {
        Map<ListNode, Integer> map = new HashMap<>();
        while (headA != null) {
            map.put(headA, headA.val);
            headA = headA.node;
        }
        while (headB !=null) {
            if (map.containsKey(headB)){
                return headB;
            }
            headB = headB.node;
        }
        return null;
    }

相交链表元素为:ListNode{val=3, node=ListNode{val=4, node=ListNode{val=5, node=null}}}
相交链表元素为:ListNode{val=3, node=ListNode{val=4, node=ListNode{val=5, node=null}}}

【LeetCode-160】相交链表_哔哩哔哩_bilibili