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

 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     int val;
 5  *     ListNode next;
 6  *     ListNode(int x) {
 7  *         val = x;
 8  *         next = null;
 9  *     }
10  * }
11  */
12 public class Solution 
13 {
14     public ListNode getIntersectionNode(ListNode headA, ListNode headB) 
15     {
16         ListNode cursor1 = headA;
17         ListNode cursor2 = headB;
18         int len1 = getListLength(cursor1);
19         int len2 = getListLength(cursor2);
20         
21         
22         if(len1 > len2)
23         {
24             for(int i=0; i<len1-len2;i++)
25                 cursor1 = cursor1.next;
26         }
27         else if(len1 < len2)
28         {
29             for(int i=0; i<len2-len1;i++)
30                 cursor2 = cursor2.next;
31         }
32         
33         while(cursor1 != null)
34         {
35             if(cursor1 == cursor2)
36                 return cursor1;
37             
38             cursor1 = cursor1.next;
39             cursor2 = cursor2.next;
40         }
41         
42         return null;
43     }
44     
45     private int getListLength(ListNode head)
46     {
47         ListNode cursor = head;
48         int len = 0;
49         
50         while(cursor != null)
51         {
52             cursor = cursor.next;
53             len++;
54         }
55         
56         return len;
57     }
58 }

参考资料:N/A

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题目来源:https://leetcode.com/

posted @ 2017-11-30 05:14  Jimmy_Cheng  阅读(177)  评论(0编辑  收藏  举报