此题扩展：链表有环，如何判断相交？
参考资料：
题目要求：
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.
Credits:
Special thanks to @stellari for adding this problem and creating all test cases.
这道题即是求两个链表交点的典型题目。具体地，我们可以这样子做：
1）求得两个链表的长度；
2）将长的链表向前移动|lenA - lenB|步；
3）两个指针一起前进，遇到相同的即是交点，如果没找到，返回nullptr。
具体程序如下：
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
if(!headA || !headB)
return nullptr; int lenA = , lenB = ;
ListNode *startA = headA, *startB = headB;
while(startA)
{
lenA++;
startA = startA->next;
} while(startB)
{
lenB++;
startB = startB->next;
} startA = headA;
startB = headB;
if(lenA > lenB)
{
for(int i = ; i < lenA - lenB; i++)
startA = startA->next;
}
else
{
for(int i = ; i < lenB - lenA; i++)
startB = startB->next;
} while(startA && startB)
{
if(startA == startB)
return startA;
startA = startA->next;
startB = startB->next;
} return nullptr;
}
};
附上该题作者分析：
There are many solutions to this problem:
- Brute-force solution (O(mn) running time, O(1) memory):
For each node ai in list A, traverse the entire list B and check if any node in list B coincides with ai.
- Hashset solution (O(n+m) running time, O(n) or O(m) memory):
Traverse list A and store the address / reference to each node in a hash set. Then check every node bi in list B: if bi appears in the hash set, then bi is the intersection node.
- Two pointer solution (O(n+m) running time, O(1) memory):
- Maintain two pointers pA and pB initialized at the head of A and B, respectively. Then let them both traverse through the lists, one node at a time.
- When pA reaches the end of a list, then redirect it to the head of B (yes, B, that's right.); similarly when pB reaches the end of a list, redirect it the head of A.
- If at any point pA meets pB, then pA/pB is the intersection node.
- To see why the above trick would work, consider the following two lists: A = {1,3,5,7,9,11} and B = {2,4,9,11}, which are intersected at node '9'. Since B.length (=4) < A.length (=6), pB would reach the end of the merged list first, because pB traverses exactly 2 nodes less than pA does. By redirecting pB to head A, and pA to head B, we now ask pB to travel exactly 2 more nodes than pA would. So in the second iteration, they are guaranteed to reach the intersection node at the same time.
- If two lists have intersection, then their last nodes must be the same one. So when pA/pB reaches the end of a list, record the last element of A/B respectively. If the two last elements are not the same one, then the two lists have no intersections.
Analysis written by @stellari.