234. Palindrome Linked List Palindrome Linked List 判断一个链表是否是回文的,很自然的想法就是两个指针,一个指针从前往后走,一个指针从后往前走,判断元素值是否相同,这里要分几个步骤来进行求解: 1、找到链表长度的一半,用追赶法,一个指针一次走两步,一个指针一次走一步 2、将后一半数组转置 3、判断链表是...
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])...
Description: Merge two sorted linked lists and return it as a new sorted list.描述:合并两个排序链表并将其作为新的排序链表返回。Hint: Use a dummy node to simplify the merge process.提示:使用虚拟节点来简化合并过程。Solution: see here 解决办法:看这里 Intersection of Two Linked Lists两个链表的...
输入:intersectVal = 2, listA = [0,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1输出:Intersected at '2'解释:相交节点的值为 2 (注意,如果两个链表相交则不能为 0)。 从各自的表头开始算起,链表 A 为 [0,9,1,2,4],链表 B 为 [3,2,4]。在 A 中,相交节点前有 3 个...
If two lists have intersection, then their last nodes must be the same one. So when pA/pB reaches the end of a list, record the last element of A/B respectively. If the two last elements are not the same one, then the two lists have no intersections. ...
Input:intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2Output:No intersectionExplanation:From the head of A, it reads as [2,6,4]. From the head of B, it reads as [1,5]. Since the two lists do not intersect, intersectVal must be 0, while skip...
LeetCode 160. Intersection of Two Linked Lists题目分析 其他 请写一个程序,找到两个单链表最开始的交叉节点。 ** 注意事项 ** 如果两个链表没有交叉,返回null。 在返回结果后,两个链表仍须保持原有的结构。 可假定整个链表结构中没有循环。 样例 下列两个链表: ...
思路1 http://bookshadow.com/weblog/2014/12/04/leetcode-intersection-two-linked-lists/ 判断交点是否存在 如果两个链表有交点,则它们的最后一个节点一定是同一个节点。所以当pA/pB到达链表末尾时,分别记录下A和B的最后一个节点。如果两个链表的末尾节点不一致,说明两个链表没有交点。
链接:https://leetcode-cn.com/problems/intersection-of-two-linked-lists python # 160.相交链表 # https://leetcode-cn.com/problems/intersection-of-two-linked-lists/solution/intersection-of-two-linked-lists-shuang-zhi-zhen-l/ class ListNode: ...
Can you solve this real interview question? Intersection of Two Linked Lists - Given the heads of two singly linked-lists headA and headB, return the node at which the two lists intersect. If the two linked lists have no intersection at all, ...