Intersection of Two Linked Lists - LeetCode
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.
思路:假设A与B长度相等,则从链表头开始一一对比即可。
假设A与B长度不等,则进行一一对比时,短的链表会先被遍历完。这时,让该指针指向另一个链表的表头,继续进行。当长链表也被遍历完时,则其指针指向短的链表。通过这种方式,一定可以找到回合点,因为两个指针遍历的总长度是相等的,都是长链表和短链表各走了一遍。
1 class Solution { 2 public: 3 ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) { 4 ListNode *ha = headA; 5 ListNode *hb = headB; 6 if (ha == NULL || hb == NULL) return NULL; 7 while (ha != NULL && hb != NULL && ha != hb) 8 { 9 ha = ha->next; 10 hb = hb->next; 11 if (ha == hb) return ha; 12 if (ha == NULL) ha = headB; 13 if (hb == NULL) hb = headA; 14 } 15 return ha; 16 } 17 };