常考问题，
先要分别遍历两个链表得到它们的长度，并求出两个长度之差。在长的链表上先遍历若干次之后，再同步遍历两个链表，知道找到相同的结点，或者一直到链表结束。此时，如果第一个链表的长度为m，第二个链表的长度为n，该方法的时间复杂度为O(m+n)
1 unsigned int GetListLength(ListNode* pHead); 2 3 ListNode* FindFirstCommonNode( ListNode *pHead1, ListNode *pHead2) 4 { 5 // 得到两个链表的长度 6 unsigned int nLength1 = GetListLength(pHead1); 7 unsigned int nLength2 = GetListLength(pHead2); 8 int nLengthDif = nLength1 - nLength2; 9 10 ListNode* pListHeadLong = pHead1; 11 ListNode* pListHeadShort = pHead2; 12 if(nLength2 > nLength1) 13 { 14 pListHeadLong = pHead2; 15 pListHeadShort = pHead1; 16 nLengthDif = nLength2 - nLength1; 17 } 18 19 // 先在长链表上走几步，再同时在两个链表上遍历 20 for(int i = 0; i < nLengthDif; ++ i) 21 pListHeadLong = pListHeadLong->m_pNext; 22 23 while((pListHeadLong != NULL) && 24 (pListHeadShort != NULL) && 25 (pListHeadLong != pListHeadShort)) 26 { 27 pListHeadLong = pListHeadLong->m_pNext; 28 pListHeadShort = pListHeadShort->m_pNext; 29 } 30 31 // 得到第一个公共结点 32 ListNode* pFisrtCommonNode = pListHeadLong; 33 34 return pFisrtCommonNode; 35 } 36 37 unsigned int GetListLength(ListNode* pHead) 38 { 39 unsigned int nLength = 0; 40 ListNode* pNode = pHead; 41 while(pNode != NULL) 42 { 43 ++ nLength; 44 pNode = pNode->m_pNext; 45 } 46 47 return nLength; 48 }