求两个链表的第一个交点

python 算法

思路：
1. 求出两个链表的长度
2. 长的链表先走两步，直至俩链表一样长
3. 依次比较俩链表的每一个值
4. 相同，直接 return 否则，return None

# -*- coding:utf-8 -*-
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
class Solution:
    def FindFirstCommonNode(self, pHead1, pHead2):
        if not pHead1 and pHead2:
            return None
        head1 = pHead1
        head2 = pHead2
        length1 = length2 = 0
        while pHead1:
            pHead1 = pHead1.next
            length1 += 1
        while pHead2:
            pHead2 = pHead2.next
            length2 += 1
        if length1 > length2:
            while length1 - length2:
                head1 = head1.next
                length1 -= 1
        else:
            while length2 - length1:
                head2 = head2.next
                length2 -= 1
        while head1 and head2:
            if head1 is head2:
                return head1
            head1 = head1.next
            head2 = head2.next
        return None