判断两个链表相交的交点

可以使用双指针法求解。首先遍历两个链表，得到它们的长度。然后让长的链表先走多出的长度，让两个链表剩下的长度相等。最后同时遍历两个链表，找到相同的节点即为它们的交点。如果没有相交的节点，就返回 null。

下面是 Java 代码实现：

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

int lenA = getLength(headA);
int lenB = getLength(headB);

ListNode curA = headA;
ListNode curB = headB;

if (lenA > lenB) {
int diff = lenA - lenB;
while (diff > 0) {
curA = curA.next;
diff--;
}
} else if (lenB > lenA) {
int diff = lenB - lenA;
while (diff > 0) {
curB = curB.next;
diff--;
}
}

while (curA != null && curB != null) {
if (curA == curB) {
return curA;
}
curA = curA.next;
curB = curB.next;
}

return null;
}

private int getLength(ListNode node) {
int count = 0;
while (node != null) {
count++;
node = node.next;
}
return count;
}
}