Leetcode|指针分别遍历双链表|剑指 Offer 52. 两个链表的第一个公共节点

class Solution {
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        if (!headA || !headB) return nullptr;
        auto nodea = headA;
        auto nodeb = headB;
        // 当两指针相遇时返回其中一个指针即可
        while (nodea != nodeb) {
            nodea = nodea ? nodea->next : headB; // 若遍历headA的指针到尾部，则转去遍历headB链表，若两链表不相交，则最后两指针均为nullptr
            nodeb = nodeb ? nodeb->next : headA;
        }
        return nodea;
    }
};

class Solution {
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        if (!headA || !headB) return nullptr;
        set<ListNode*> s;
        auto nodea = headA;
        auto nodeb = headB;
        while (nodea || nodeb) {
            if (nodea) {
                if (s.count(nodea)) return nodea;
                else s.insert(nodea);
                nodea = nodea->next;
            }
            if (nodeb) {
                if (s.count(nodeb)) return nodeb;
                else s.insert(nodeb);
                nodeb = nodeb->next;
            }
        } 
        return nullptr;
    }
};