代码:
class Solution {
public:
ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
ListNode *pa = headA, *pb = headB;
while ( pa != pb) {
pa = (pa != NULL)? pa->next: headB;
pb = (pb != NULL)? pb->next: headA;
}
return pa;
}
};
时间复杂度:O(m+n)。
空间复杂度:O(1)。



