380. Intersection of Two Linked Lists【medium】

最新推荐文章于 2024-07-02 12:57:50 发布

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最新推荐文章于 2024-07-02 12:57:50 发布

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原文链接：http://www.cnblogs.com/abc-begin/p/8157802.html

本文提供了一种在两个单链表中找到它们开始相交节点的方法。通过计算两个链表的长度并调整较长链表的起始位置，使得两链表剩余部分长度相等，然后同时遍历两个链表直至找到相同的节点。

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Write a program to find the node at which the intersection of two singly linked lists begins.

Notice

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.

Example

The following two linked lists:

A:          a1 → a2
                   ↘
                     c1 → c2 → c3
                   ↗            
B:     b1 → b2 → b3

begin to intersect at node c1.

Challenge

Your code should preferably run in O(n) time and use only O(1) memory.

解法一：

 1 class Solution {
 2 public:
 3     /**
 4      * @param headA: the first list
 5      * @param headB: the second list
 6      * @return: a ListNode
 7      */
 8     ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
 9         // write your code here
10         if(headA == NULL || headB == NULL)
11             return NULL;
12         ListNode* iter1 = headA;
13         ListNode* iter2 = headB;
14         int len1 = 1;
15         while(iter1->next != NULL)
16         {
17             iter1 = iter1->next;
18             len1 ++;
19         }
20         int len2 = 1;
21         while(iter2->next != NULL)
22         {
23             iter2 = iter2->next;
24             len2 ++;
25         }
26         if(iter1 != iter2)
27             return NULL;
28         if(len1 > len2)
29         {
30             for(int i = 0; i < len1-len2; i ++)
31                 headA = headA->next;
32         }
33         else if(len2 > len1)
34         {
35             for(int i = 0; i < len2-len1; i ++)
36                 headB = headB->next;
37         }
38         while(headA != headB)
39         {
40             headA = headA->next;
41             headB = headB->next;
42         }
43         return headA;
44     }
45 };