《剑指offer》两个链表中的第一个公共节点

本题来自《剑指offer》 两个链表中的第一个公共节点

题目：

 　　输入两个链表，找出它们的第一个公共结点。

思路：

 　　交叉链表形式像Y形，这样两个分叉长短不一致，首先分别统计各自的长短，分叉为5和4那么，相差1个节点，让多的那个先走1个节点，之后同时遍历，当值相等即是公共节点。

Python Code：

# -*- 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 or not pHead2:                        #empty node judgement
            return None
        m = 0                                               #the length of list phead1
        M = pHead1
        while M.next:                                       #loop list phead1
            m += 1
            M = M.next                                      #the next node of phead1
        n = 0                                               #the lengteh of list phead2
        N = pHead2
        while N.next:                                       #loop list phead2
            n += 1
            N = N.next                                      #the next node of phead2
        if m > n:                                           #if list phead1 is longer than list phead2
            for i in range(m - n):                          #traversing m-n node in advance
                pHead1 = pHead1.next                        #the next node of phead1
        elif n > m:                                         #same as above
            for i in range(n - m):
                pHead2 = pHead2.next
        else :
            pass
        while pHead1 and pHead2:                            #traverse at the same time
            if pHead1.val == pHead2.val:                    #during this process,only when the two value is equal
                return pHead1                               #return one of them
            elif not pHead1.next or not pHead2.next:        #when there is no equal value when travesing to the end
                return None                                 #there are no public node
            else:
                pHead1 = pHead1.next                        #the next node of the list phead1
                pHead2 = pHead2.next                        #same as above
                

总结：