力扣HOT100 - 160. 相交链表
解题思路：
/**
* 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) {
if (headA == null || headB == null) return null;
ListNode pa = headA;
ListNode pb = headB;
while (pa != pb) {
pa = (pa != null) ? pa.next : headB;
pb = (pb != null) ? pb.next : headA;
if (pa == null && pb == null) return null;
}
return pa;
}
}原文地址：https://blog.csdn.net/qq_61504864/article/details/137633400
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