寻找两链表中的公共点的办法:

定义两个指针N0,N1分别从两个链表头开始遍历,将一个链表遍历完之后指向另一个链表头,则有如下情况:

1.若两个链表有公共点,则N0=N1时,正好处于两个链表的公共节点位置;

2.若两个链表没有公共点,于遍历的速度是相等的,无论链表的规模如何,此时N0和N1都为None,N0=N1也成立;

3.需要判断pHead1或pHead2为零的边界条件。

class Solution:
    def FindFirstCommonNode(self , pHead1 , pHead2 ):
        # write code here
        N1 = pHead1
        N2 = pHead2
        if N1 is None or N2 is None:
            return None
        i = 0
        j = 0
        while N1 != N2:
            N1 = N1.next 
            N2 = N2.next 
            if N1 is None and i < 1:
                i += 1
                N1 = pHead2
            if N2 is None and j < 1:
                j += 1
                N2 = pHead1
        
        if N1 is None:
            return None
        else:
            return N1