LeetCode 160. Intersection of Two Linked Lists （两个链表的交点）

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.

---

题目标签：Linked List

　　题目给了我们两个链表，让我们找到它们是否有交点，返回那个交点。

　　试想一下，如果是两个长度相等的链表，那么我们只需要遍历链表，比较两个点 是否 相等 就可以了。

　　所以，我们首先要得到两个链表的长度，如果不一样长，那么把长的链表先走，走到 和 另外一个链表一样长度的时候，开始比较两个点 是否 相等来找到交点；如果没有，那么最后就返回null。

　　如果两个链表有交点的话，那么从交点前一个点开始，两个链表从这里开始，之后的长度一定是相等的。

Java Solution:

Runtime beats 41.31%

完成日期：06/09/2017

关键词：singly-linked list

关键点：让更长的链表先走完多余的部分

 /**
  * Definition for singly-linked list.
  * public class ListNode {
  *     int val;
  *     ListNode next;
  *     ListNode(int x) {
  *         val = x;
  *         next = null;
  *     }
  * }
  */
 public class Solution
 {
     public ListNode getIntersectionNode(ListNode headA, ListNode headB)
     {
         ListNode cursor1 = headA;
         ListNode cursor2 = headB;
         int len1 = getListLength(cursor1);
         int len2 = getListLength(cursor2);

         if(len1 > len2)
         {
             for(int i=0; i<len1-len2;i++)
                 cursor1 = cursor1.next;
         }
         else if(len1 < len2)
         {
             for(int i=0; i<len2-len1;i++)
                 cursor2 = cursor2.next;
         }

         while(cursor1 != null)
         {
             if(cursor1 == cursor2)
                 return cursor1;

             cursor1 = cursor1.next;
             cursor2 = cursor2.next;
         }

         return null;
     }

     private int getListLength(ListNode head)
     {
         ListNode cursor = head;
         int len = 0;

         while(cursor != null)
         {
             cursor = cursor.next;
             len++;
         }

         return len;
     }
 }

参考资料：N/A

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