【牛客题霸题解 】两个链表的第一个公共结点

两个链表的长度如果相等的话，从头节点开始走，两个指针相等的时候就找到了第一个公共节点。
如果两个链表的长度不相等，假设 比 多 个结点，那就从 的第 个结点开始走， 从头结点开始走，，两个指针相等的时候就找到了第一个公共节点。
然后问题就转化为了如何知道哪个链表的长度更长一点，比长度短的链表多了几个结点。解决这个问题的话，我们只需要先设置两个指针，分别指向两个链表的头节点，然后当一个指针指向空了以后开始计数，看另一个指针走多少步才会指向空。

c++

class Solution {
public:
    ListNode* FindFirstCommonNode( ListNode* pHead1, ListNode* pHead2) {
        ListNode* L1 = pHead1;
        ListNode* L2 = pHead2;
        int l1 = 0,l2 = 0;
        while(L1 !=NULL || L2!= NULL)
        {
            if(L1!=NULL){L1 = L1->next;}
            else{ l2++; }
            if(L2 != NULL){L2 = L2->next;}
            else{ l1++; }
        }
        L1 = pHead1;
        L2 = pHead2;
        while(l1!=0)
        {
            L1 = L1->next;
            l1--;
        }
        while(l2!=0)
        {
            L2 = L2->next;
            l2--;
        }
        while(L1!=L2)
        {
            L1 = L1->next;
            L2 = L2->next;
        }
        return L1;
    }
};

java

public class Solution {
    public ListNode FindFirstCommonNode(ListNode pHead1, ListNode pHead2) {
        ListNode L1 = pHead1;
        ListNode L2 = pHead2;
        int l1 = 0,l2 = 0;
        while(L1 != null || L2!= null)
        {
            if(L1!= null){L1 = L1.next;}
            else{ l2++; }
            if(L2 != null){L2 = L2.next;}
            else{ l1++; }
        }
        L1 = pHead1;
        L2 = pHead2;
        while(l1!=0)
        {
            L1 = L1.next;
            l1--;
        }
        while(l2!=0)
        {
            L2 = L2.next;
            l2--;
        }
        while(L1!=L2)
        {
            L1 = L1.next;
            L2 = L2.next;
        }
        return L1;
    }
}

python

class Solution:
    def FindFirstCommonNode(self , pHead1 , pHead2 ):
        # write code here
        L1 = pHead1
        L2 = pHead2
        l1 = 0
        l2 = 0
        while L1 != None or L2!= None:
            if L1!= None:
                L1 = L1.next
            else:
                l2 = l2+1
            if L2 != None:
                L2 = L2.next
            else:
                l1 = l1+1
        L1 = pHead1
        L2 = pHead2
        while l1!=0:
            L1 = L1.next
            l1 = l1-1
        while l2!=0:
            L2 = L2.next
            l2 = l2-1
        while L1!=L2:
            L1 = L1.next
            L2 = L2.next
        return L1