《剑指offer》——两个链表的第一个公共结点

时间复杂度为O(m+n)。

struct ListNode{    int val;    struct ListNode *next;    ListNode(int x) :        val(x), next(NULL){}};ListNode* FindFirstCommonNode( ListNode *pHead1, ListNode *pHead2) {    ListNode* pFirstCommonNode = NULL;    if(pHead1 == NULL || pHead2 == NULL)//若输入链表有一个为空，返回空结点        return pFirstCommonNode;    ListNode* pNode1 = pHead1;    ListNode* pNode2 = pHead2;    int pLen1 = 0, pLen2 = 0;    while(pNode1)//分别计算两个链表的长度    {        pNode1 = pNode1 -> next;        pLen1++;    }    while(pNode2)    {        pNode2 = pNode2 -> next;        pLen2++;    }    int offSet = 0;    if(pLen1 < pLen2)//移动链表的头结点，将长链表的头结点与短链表的头结点对齐    {        offSet = pLen2 - pLen1;        while(offSet)        {            pHead2 = pHead2 -> next;            offSet--;        }    }    else if(pLen2 < pLen1)    {        offSet = pLen1 - pLen2;        while(offSet)        {            pHead1 = pHead1 -> next;            offSet--;        }    }    while(pHead1 && pHead2)//如果当前结点不相等，则继续同步遍历两个链表直到找到第一个公共结点    {        if(pHead1 == pHead2)        {            pFirstCommonNode = pHead1;            break;        }        else        {            pHead1 = pHead1 -> next;            pHead2 = pHead2 -> next;        }    }    return pFirstCommonNode;//返回第一个公共结点}