while headB: listB.append(headB.val) headB=headB.next minlen=len(listA) if len(listA)<len(listB) else len(listB) print listA,listB,minlen if listA[-1]!=listB[-1]:return None for i in xrange(1,minlen+1): print i if listA[-i]!=listB[-i]: return ListNode(listA[-i+1]) if i==minlen: return ListNode(listA...
Maintain two pointers pA and pB initialized at the head of A and B, respectively. Then let them both traverse through the lists, one node at a time. When pA reaches the end of a list, then redirect it to the head of B (yes, B, that's right.); similarly when pB reaches the end...
15. 用set来把list的重复元素过滤掉,然后判断是否存在,把结果保存起来 http://www.waitingfy.com/archives/3724
Suppose we have two sorted linked lists L1 and L2, we have to make a new sorted linked list which contains the intersection of these two lists. So, if the input is like L1 = [2, 4, 8] L2 = [3, 4, 8, 10], then the output will be [4, 8, ] To solve this, we will ...
Learn how to find the intersection of two arrays in Python with our step-by-step guide and example code.
python 内置的linkedlist python list intersection 1. 什么是列表 列表是一个可变的数据类型 列表由[]来表示, 每一项元素使用逗号隔开. 列表什么都能装. 能装对象的对象. 列表可以装大量的数据 2. 列表的索引和切片 列表和字符串一样. 也有索引和切片. 只不过切出来的内容是列表...
LeetCode笔记:160. Intersection of Two Linked Lists 问题: Write a program to find the node at which the intersection of two singly linked lists begins. For example, the following two linked lists: image.png begin to intersect at node c1. Notes:...
If the two linked lists have no intersection at all, return null.d 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)...
350. Intersection of Two Arrays II Given two arrays, write a function to compute their intersection. Example 1: Example 2: Note: Each element in the result should appear as many times as it shows in both arrays. The result can be in an......
题目链接https://leetcode.com/problems/intersection-of-two-linked-lists/ 题意很好懂,问题在于如何找到相交的node,想到的最简单的方法从node的最后一个节点去往前数,遇到分叉,则返回当前节点,否则返回None。这里用 两个list去保存node,从list的最后一个位置开始往前进。 代码如下,提交后,超过 99.84% 提交代码的...