Write a program to find the node at which the intersection of two singly linked lists begins.
For example, the following two linked lists:
A: a1 → a2
↘
c1 → c2 → c3
↗
B: b1 → b2 → b3
begin to intersect at node c1.
Notes:
- If the two linked lists have no intersection at all, return
null. - The linked lists must retain their original structure after the function returns.
- You may assume there are no cycles anywhere in the entire linked structure.
- Your code should preferably run in O(n) time and use only O(1) memory.
class Solution {
public:
ListNode *getIntersectionNode(ListNode *headA, ListNode *headB)
{
ListNode *p1 = headA;
ListNode *p2 = headB;
if (p1 == NULL || p2 == NULL) return NULL;
while (p1 != NULL && p2 != NULL && p1 != p2)
{
p1 = p1->next;
p2 = p2->next;
if (p1 == p2) return p1; //当达到相遇点时,直接返回p1或者p2
if (p1 == NULL) p1 = headB; //当p1比p2先到达结尾时,重新将其移动到第二个链表的头部
if (p2 == NULL) p2 = headA; //当p2比p1先到达结尾时,重新将其移动到第二个链表的头部,这样的话,二者到相遇点的距离是相等的
}
return p1;
}
};

本文介绍了一种高效算法,用于在两个单链表中找到它们相交的节点。通过移动两个指针并调整它们的位置,直到它们在相交点相遇。该方法在O(n)时间内运行,并且仅使用O(1)内存。
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