两个节点的第一个公共节点

思路：既然是公共享有的节点，必然最后公共节点同即结尾是一样的（公共节点），找到最短的链表，长链表移动到和短链表一样长的位置，同时遍历，判断即找到。

/*
struct ListNode {
	int val;
	struct ListNode *next;
	ListNode(int x) :
			val(x), next(NULL) {
	}
};*/
class Solution {
public:
    ListNode* FindFirstCommonNode( ListNode* pHead1, ListNode* pHead2) {
        int n = ListLength(pHead1),m = ListLength(pHead2);
        int k = abs(n-m);
        if(n > m){
            while(k){
                pHead1 = pHead1->next;
                --k;
            }
        }else{
            while(k){
                pHead2 = pHead2->next;
                --k;
            }
        }
        while(pHead1 && pHead2){
            if(pHead1->val == pHead2->val)
                 return pHead1;
             pHead1 = pHead1->next;
             pHead2 = pHead2->next;
        }
        return pHead1;
    }
    int ListLength(ListNode* head){
        ListNode* h = head;
        int len = 0;
        while(h){
            ++len;
            h = h->next;
        }
        return len;
    }
};