LeetCode之“链表”：Intersection of Two Linked Lists

　　此题扩展：链表有环，如何判断相交？

　　参考资料：

　　　　编程判断两个链表是否相交

　　　　面试精选：链表问题集锦

　　题目链接

　　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.