题目

intersection-of-two-linked-lists

算法

* 相遇算法

代码

* 相遇算法

class Solution {
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        if (!headA || !headB) return NULL;
        ListNode *a = headA, *b = headB;
        while (a != b) {
            a = a ? a->next : headB;
            b = b ? b->next : headA;
        }
        return a;
    }
};