腾讯五十题 No.35 相交链表

题目链接

public class Solution {
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        if(headA == null || headB == null) return null;
        ListNode pa = headA,pb = headB;
        while(pa != pb){
            pa  = pa != null? pa.next : headB;
            pb  = pb != null? pb.next : headA;
        }
        return pa;
    }
}

标签：No.35,pb,ListNode,链表,pa,headB,headA,腾讯,null
来源： https://www.cnblogs.com/jianjiana/p/15873487.html