剑指offer（33）-两个链表的第一个公共结点

题目描述

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输入两个链表，找出它们的第一个公共结点。

　代码

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/*struct ListNode {    int val;    struct ListNode *next;    ListNode(int x) :            val(x), next(NULL) {    }};*/class Solution {public:    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->next;    }    while((pListHeadLong != NULL) && (pListHeadshort != NULL) && (pListHeadLong != pListHeadshort)) {        pListHeadshort = pListHeadshort->next;        pListHeadLong = pListHeadLong->next;    }    // 得到第一个公共节点    ListNode *pFirstCommonNode = pListHeadLong;    return pFirstCommonNode;    }    // 获取链表的长度unsigned int GetListLength(ListNode *pHead){    unsigned int nLength = 0;    ListNode *pNode = pHead;    while(pNode != NULL) {        nLength ++;        pNode = pNode->next;    }    return nLength;}};