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找到第一个相同的节点

让长的链表的指针先移动差值距离，然后同时移动，判断是否当前节点相同

public class Solution {    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {        int sizeA = 0;        int sizeB = 0;        ListNode curA = headA;        ListNode curB = headB;        while (curA != null) {            sizeA++;            curA = curA.next;        }        while (curB != null) {            sizeB++;            curB = curB.next;        }        while (sizeA > sizeB) {            headA = headA.next;            sizeA--;        }        while (sizeA < sizeB) {            headB = headB.next;            sizeB--;        }        while (headA != headB) {            headA = headA.next;            headB = headB.next;        }        return headA;    }}

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