两个链表的第一个公共结点

题目：输入两个链表，找出它们的第一个公共结点。链表结点定义如下：struct ListNode{    int       m_nkey;    ListNode* m_pNext;};

分析：如果两个单链表有公共结点，那么公共结点之后的结点全都是是两个链表的公共结点。我们可以从后往前遍历两个单链表，直到遍历到的结点不同为止。
一、使用栈可以实现单链表的反向遍历。

class Solution {public:    ListNode* FindFirstCommonNode( ListNode* pHead1, ListNode* pHead2 )    {        stack<ListNode*> list1;        stack<ListNode*> list2;        ListNode* pList = NULL;        while ( pHead1 != NULL )        {            list1.push( pHead1 );            pHead1 = pHead1->m_pNext;        }        while ( pHead2 != NULL )        {            list2.push( pHead2 );            pHead2 = pHead2->m_pNext;        }        while ( list1.empty() || list2.empty() )        {            if ( list1.top() != list2.top() )            {                return pList;            }            pList = list1.top();            list1.pop();        }        return pList;     }};

二、先将两个链表遍历一遍，得到两个链表的长度，然后先在较长链表走几步，然后同时前进，这种方法的时间复杂度也是O(m+n)，但是不需要额外的辅助空间。

class Solution {public:    ListNode* FindFirstCommonNode( ListNode* pHead1, ListNode* pHead2) {        if ( pHead1 == NULL || pHead2 == NULL )            return NULL;        int lenList1, lenList2;        ListNode* ppHead1 = pHead1;        ListNode* ppHead2 = pHead2;        while ( pHead1 != NULL )        {            lenList1++;            pHead1 = pHead1->next;        }        while ( pHead2 != NULL )        {            lenList2++;            pHead2 = pHead2->next;        }        pHead2 = ppHead2;        pHead1 = ppHead1;        int forward = lenList1 - lenList2;        if ( forward < 0 )        {            forward = -forward;            while ( forward-- )            {                pHead2 = pHead2->next;            }        }        else         {            while ( forward-- )            {                pHead1 = pHead1->next;            }        }        while ( pHead1 != pHead2 && pHead1 != NULL && pHead2 != NULL )        {            pHead1 = pHead1->next;            pHead2 = pHead2->next;        }        return pHead1;    }};