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, returnnull.
• 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 1: 先把2 list移动到等长位置，再同时移动，直到找到交点

    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        if(headA == null || headB == null){
            return null;
        }

        int l1 = findLength(headA), l2 = findLength(headB);
        if(l1 < l2){
            ListNode temp = headA;
            headA = headB;
            headB = temp;
        }
        for(int i = 0; i < Math.abs(l1 - l2); i++){
            headA = headA.next;
        }
        if(headA == headB){
            return headA;
        }
        while(headA != null){
            if(headA.next == headB.next){
                return headA.next;
            }
            headA = headA.next;
            headB = headB.next;
        }
        return null;
    }
    private int findLength(ListNode head){
        int length = 0;
        while(head != null){
            length++;
            head = head.next;
        }
        return length;
    }

Solution 2:

1) 找到第一个Linked List的最后一个node， 将其和第二个Linked List的head相连

2) 若两个Linked List相交，则一定会形成一个环（若无环则证明不相交，返回null）

3) 再以第一个Linked List的head为起点寻找环的起点即可

4) 为保持两个Linked List结构，最后要将第一个Linked List的最后一个节点指向null