CodingInterview-两个链表的第一个公共结点

思路比较简单，先找出链表的长度，接着将长的链表先走一段长度，然后长链表和短链表一起继续走，直到遇到相同的节点。

/*
struct ListNode {
	int val;
	struct ListNode *next;
	ListNode(int x) :
			val(x), next(NULL) {
	}
};*/
class Solution {
public:
    ListNode* FindFirstCommonNode( ListNode* pHead1, ListNode* pHead2) {
        if(pHead1==NULL || pHead2==NULL) return NULL;
        int len1=GetListLength(pHead1);
        int len2=GetListLength(pHead2);
        ListNode* pLonger=pHead1;
        ListNode* pShorter=pHead2;
        if(len1<len2){
            pLonger=pHead2;
            pShorter=pHead1;
        }
        int lenDiff=len1>len2?len1-len2:len2-len1;
        for(int i=0;i<lenDiff;i++){
            pLonger=pLonger->next;
        }
        while(pLonger!=NULL && pShorter!=NULL){
            if(pLonger==pShorter) return pLonger;
            pLonger=pLonger->next;
            pShorter=pShorter->next;
        }
        return NULL;
    }
    int GetListLength(ListNode* pHead){
        ListNode* p=pHead;
        int len=0;
        while(p!=NULL){
            len++;
            p=p->next;
        }
        return len;
    }
};