面试题37：两个链表的第一个公共结点

题目描述

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：

题目分析

剑指Offer（纪念版）P193 思路三

代码实现

ListNode* FindFirstCommonNode( ListNode *pHead1, ListNode *pHead2)
{
    // 得到两个链表的长度
    unsigned int nLength1 = GetListLength(pHead1);
    unsigned int nLength2 = GetListLength(pHead2);
    int nLengthDif = nLength1 - nLength2;
 
    ListNode* pListHeadLong = pHead1;
    ListNode* pListHeadShort = pHead2;
    if(nLength2 > nLength1)
    {
        pListHeadLong = pHead2;
        pListHeadShort = pHead1;
        nLengthDif = nLength2 - nLength1;
    }
 
    // 先在长链表上走几步，再同时在两个链表上遍历
    for(int i = 0; i < nLengthDif; ++ i)
        pListHeadLong = pListHeadLong->m_pNext;
 
    while((pListHeadLong != NULL) && 
        (pListHeadShort != NULL) &&
        (pListHeadLong != pListHeadShort))
    {
        pListHeadLong = pListHeadLong->m_pNext;
        pListHeadShort = pListHeadShort->m_pNext;
    }
 
    // 得到第一个公共结点
    ListNode* pFisrtCommonNode = pListHeadLong;
 
    return pFisrtCommonNode;
}

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