面试题52：两个链表的第一个公共节点

一、题目

    输入两个链表，找出它们的第一个公共节点。

二、解法

    思路：假设链表1比链表2长。 则从两个链表的头结点开始遍历，前提是先让较长的链表前进（length1-length2）个节点，然后同时开始遍历，它们可以同时到达第一个公共节点。

/*
struct ListNode {
	int val;
	struct ListNode *next;
	ListNode(int x) :
			val(x), next(NULL) {
	}
};*/
class Solution {
public:
    unsigned int GetListLength(ListNode* pHead)
    {
        unsigned int length = 0;
        ListNode* pNode = pHead;
        while(pNode!=nullptr)
        {
            ++length;
            pNode = pNode->next;
        }
        return length;
    }
    ListNode* FindFirstCommonNode( ListNode* pHead1, ListNode* pHead2) {
        unsigned int length1 = GetListLength(pHead1);
        unsigned int length2 = GetListLength(pHead2);
        int dist = length1 - length2;
        
        ListNode* pListHeadLong = pHead1;
        ListNode* pListHeadShort = pHead2;
        if(length2>length1)
        {
            pListHeadLong = pHead2;
            pListHeadShort = pHead1;
            dist = length2-length1;
        }
        for(int i=0; i<dist; ++i)
            pListHeadLong = pListHeadLong->next;
        while(pListHeadLong!=nullptr && pListHeadShort!=nullptr && pListHeadLong!=pListHeadShort)
        {
            pListHeadLong = pListHeadLong->next;
            pListHeadShort = pListHeadShort->next;
        }
        return pListHeadLong;
    }
};