剑指Offer 36. 两个链表的第一个公共结点 （链表）

题目描述

输入两个链表，找出它们的第一个公共结点。

题目地址

https://www.nowcoder.com/practice/6ab1d9a29e88450685099d45c9e31e46?tpId=13&tqId=11189&tPage=2&rp=2&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking

思路

思路1：把两个链表拼接起来，一个pHead1在前pHead2在后，一个pHead2在前pHead1在后，这样，生成了两个相同长度的链表，我们只要同时遍历这两个表，就一定能找到公共节点。时间复杂度O(m+n)，空间复杂度O(m+n).

思路2：首先依次遍历两个链表，记录两个链表的长度m和n，如果 m > n，那么我们就先让长度为m的链表走m-n个结点，然后两个链表同时遍历，当遍历到相同的结点的时候停止即可。对于 m < n，同理。

Python

# -*- coding:utf-8 -*-
class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None
node1 = ListNode(1)
node2 = ListNode(2)
node3 = ListNode(3)
node4 = ListNode(4)
node5 = ListNode(5)
node6 = ListNode(6)
node1.next = node2
node2.next = node5
node5.next = node6
node3.next = node4
node4.next = node5
class Solution:
    def FindFirstCommonNode(self, pHead1, pHead2):
        # 方法1
        # if not pHead1 or not pHead2:
        #     return None
        # cur1, cur2 = pHead1, pHead2
        # while cur1 != cur2:
        #     cur1 = cur1.next if cur1 else pHead2
        #     cur2 = cur2.next if cur2 else pHead1
        # return cur1

        # 方法2
        if not pHead1 or not pHead2:
            return None
        m = n = 0
        cur1, cur2 = pHead1, pHead2
        while cur1:
            m += 1
            cur1 = cur1.next
        while cur2:
            n += 1
            cur2 = cur2.next
        if m > n:
            for i in range(m-n):
                pHead1 = pHead1.next
            while pHead1:
                if pHead1 == pHead2:
                    return pHead1
                pHead1, pHead2 = pHead1.next, pHead2.next
        else:
            for i in range(n-m):
                pHead2 = pHead2.next
            while pHead1:
                if pHead1 == pHead2:
                    return pHead1
                pHead1, pHead2 = pHead1.next, pHead2.next
        return None

if __name__ == '__main__':
    result = Solution().FindFirstCommonNode(node1, node3)
    while result:
        print(result.val, end = ' ')
        result = result.next