题解 | #两个链表的第一个公共结点#

解题思路：

1. 两个链表有长有短，先把长短确定下来
2. 短的链表不动，长的链表往后面移，移它们的长度差
3. 为什么要长的往后移呢？ - 两个链表的的公共节点，顶天把短的整个作为公共链了，所以长链往后移
4. 由于返回值需要的是一个链表，所以开辟一个新链，找到一个公共部分后链接上就行了

#     def __init__(self, x):
#         self.val = x
#         self.next = None

#
# 
# @param pHead1 ListNode类 
# @param pHead2 ListNode类 
# @return ListNode类
#
class Solution:
    def FindFirstCommonNode(self , pHead1 , pHead2 ):
        # write code here
        if not pHead1 or not pHead2:
            return None
        res = ListNode(0)
        r = res
        p = pHead1
        q = pHead2
        count1 = 0
        count2 = 0
        while p:
            p = p.next
            count1 += 1
        while q:
            q = q.next
            count2 += 1
        p, q = pHead1, pHead2
        if count1 > count2:  # # pHead1 > pHead2
            for i in range(count1-count2):
                p = p.next
        else:
            for i in range(count2-count1):
                q = q.next
        while p and q:
            if p.val == q.val:
                r.next = p
                r = r.next
            p = p.next
            q = q.next
        return res.next