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。

#include <iostream>
using namespace std;

ListNode
{
 int m_data;
 ListNode *m_next;
};

unsigned int GetListLength(ListNode* pHead)
{
 unsigned int nLength=0;
 ListNode* pNode=pHead;
 while(pNode!=NULL)
 {
  ++nLength;
  pNode=pNode->m_next;
 }
}

ListNode* FindFirstCommonNode(ListNode* pHead1,ListNode* pHead2)
{
 unsigned int nLength1=GetListLength(pHead1);
 unsigned int nLength2=GetListLength(pHead2);
 int nLengthDif=nLength1-nLength2;
 ListNode* pListLong=pHead1;
 ListNode* pListShort=pHead2;
 if (nLength2>nLength1)
 {
  pListLong=pHead2;
  pListShort=pHead1;
  nLengthDif=-nLengthDif;
 }
 for (int i=0;i<nLengthDif;i++)
 {
  pListLong=pListLong->m_next;
 }
 while((pListLong!=NULL)&&(pListShort!=NULL)&&(pListLong!=pListShort))
 {
  pListLong=pListLong->m_next;
  pListShort=pListShort->m_next;
 }
 ListNode* pFirstCommonNode=pListLong;
 return pFirstCommonNode;
}