题目: 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 lis…

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…

/** * Definition for singly-linked list. * public class ListNode { * public int val; * public ListNode next; * public ListNode(int x) { val = x; } * } */ public class Solution { public ListNode GetIntersectionNode(ListNode headA, ListNode headB) { if…

编写一个程序,找到两个单链表相交的起始节点. 例如,下面的两个链表: A: a1 → a2 ↘ c1 → c2 → c3 ↗ B: b1 → b2 → b3 在节点 c1 开始相交. 注意: 如果两个链表没有交点,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. 程序尽量满足 O(n) 时间复杂度,且仅用 O(1) 内存. /** * Definition for singly-linked list. * public class ListNode…

题目描述: 方法一:动态规划 class Solution: def f(self, n, m): if n < m: n, m = m, n if (n, m) in self.mem: return self.mem[(n, m)] if n == m: ans = else: ans = n*m , n): ans = min(ans, self.f(i, m) + self.f(n-i, m)) , m): ans = min(ans, self.f(n, i) + self.f(n,…

题目描述: 自己的提交:O(2**n∗n∗m),m 为字符串长度 class Solution: def maxLength(self, arr: List[str]) -> int: from collections import Counter res = set() n = len(arr) def helper(idx, tmp): : return count = Counter(tmp) for i in count.values()): res.add(tmp) for i in…

题目描述: 参考格雷编码: class Solution: def circularPermutation(self, n: int, start: int) -> List[int]: res = [] ** n): res.append((i >> ) ^ i) start = res.index(start) return res[start:] + res[:start]…

题目描述: class Solution: def findSolution(self, customfunction: 'CustomFunction', z: int) -> List[List[int]]: res = [] ,): ,): if customfunction.f(i,j) == z: res.append([i,j]) if customfunction.f(i,j) > z: break return res 另:O(N) class Solution(object)…

题目: click here!!题目传送门 思路: 1.笨方法 因为如果两个链表相交的话,从相交的地方往后是同一条链表,所以: 分别遍历两个链表,得出两个链表的长度,两个长度做差得到n,然后将长的链表头指针先移动n个结点,然后两个链表再同时移动,如果出现两个链表的指针直到同一个内存地址,说明相交,没有出现指向同一个内存地址的情况就是不相交 class Solution { public ListNode getIntersectionNode(ListNode headA, ListNode h…

朋友遇到过的t面试题 leetcode160 找链表交点 leetcode206 反转链表…