剑指offer——两个链表的第一个公共结点

首先计算两个链表的长度m，n，让较长的链表先走|m-n|，然后两个链表一起走，直到找到元素相同的结点。

 1 #include <iostream>
 2 using namespace std;
 3 
 4 ListNode
 5 {
 6     int m_data;
 7     ListNode *m_next;
 8 };
 9 
10 unsigned int GetListLength(ListNode* pHead)
11 {
12     unsigned int nLength=0;
13     ListNode* pNode=pHead;
14     while(pNode!=NULL)
15     {
16         ++nLength;
17         pNode=pNode->m_next;
18     }
19 }
20 
21 ListNode* FindFirstCommonNode(ListNode* pHead1,ListNode* pHead2)
22 {
23     unsigned int nLength1=GetListLength(pHead1);
24     unsigned int nLength2=GetListLength(pHead2);
25     int nLengthDif=nLength1-nLength2;
26     ListNode* pListLong=pHead1;
27     ListNode* pListShort=pHead2;
28     if (nLength2>nLength1)
29     {
30         pListLong=pHead2;
31         pListShort=pHead1;
32         nLengthDif=-nLengthDif;
33     }
34     for (int i=0;i<nLengthDif;i++)
35     {
36         pListLong=pListLong->m_next;
37     }
38     while((pListLong!=NULL)&&(pListShort!=NULL)&&(pListLong!=pListShort))
39     {
40         pListLong=pListLong->m_next;
41         pListShort=pListShort->m_next;
42     }
43     ListNode* pFirstCommonNode=pListLong;
44     return pFirstCommonNode;
45 }

 



 

posted @ 2014-09-09 15:02  啵啵那个臭  阅读(152)  评论(0)  编辑  收藏  举报