(LeetCode)Intersection of Two Linked Lists --- 两个链表的交集

Write a program to find the node at which the intersection of two singly linked lists begins.

For example, the following two linked lists:

A:          a1 → a2                   ↘                     c1 → c2 → c3                   ↗            B:     b1 → b2 → b3

begin to intersect at node c1.

Notes:

• If the two linked lists have no intersection at all, return null.
• The linked lists must retain their original structure after the function returns.
• You may assume there are no cycles anywhere in the entire linked structure.
• Your code should preferably run in O(n) time and use only O(1) memory.

Credits:
Special thanks to @stellari for adding this problem and creating all test cases.

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解题分析：

对于这个题目来说，我的做法就是先判断出哪个LinkedList 比较小。

然后可循环比较最小的次数，来达到最高的效率，没有环，更好判断一些。

加上条件判断，就可以判断出来是否末尾的串有交集。

# -*- coding:utf-8 -*-__author__ = 'jiuzhang'class ListNode(object):    def __init__(self, x):        self.val = x        self.next = Noneclass Solution(object):    def getIntersectionNode(self, headA, headB):        """        :type head1, head1: ListNode        :rtype: ListNode        """        listA= []        listB= []        if headA == None or headB == None:            return None        while 1:            if headA == None:                break            listA.append(headA.val)            headA = headA.next        while 1:            if headB == None:                break            listB.append(headB.val)            headB = headB.next        if listA[-1] != listB[-1]:            return None        if len(listA) < len(listB):            minLen = len(listA)        else:            minLen = len(listB)        for i in range(1, minLen + 1):            if listA[-i] != listB[-i]:                return ListNode(listA[-i + 1])            if i == minLen:                return ListNode(listA[-i])