141. Linked List Cycle(Easy)2019.7.10 题目地址https://leetcode.com/problems/linked-list-cycle/ Copy Given a linked list, determineifit has a cycleinit.Torepresent a cycleinthe given linked list, we use anintegerpos which represents the position (0-indexed)inthe linked listwheretail connectst...
https://leetcode-cn.com/problems/set-mismatch/solution/cuo-wu-de-ji-he-by-leetcode/
LinkedList Cycle (easy) Start of LinkedList Cycle (medium) Happy Number (medium) Middle of the LinkedList (easy) 4. Pattern: Merge Intervals,区间合并类型 区间合并模式是一个用来处理有区间重叠的很高效的技术。在设计到区间的很多问题中,通常咱们需要要么判断是否有重叠,要么合并区间,如果他们重叠的话。这...
:return: list[list[int]] """stack=[(root,0)]res=[]# 如果栈不为空whilestack:# 将栈中最后一个元素弹出node,level=stack.pop()# 如果该结点存在ifnode:# 如果结果列表的长度小于层数+1iflen(res)<level+1:res.insert(0,[])# 将当前结点的值添加在结果列表中res[-(level+1)].append(node.val...
Reverse a LinkedList (easy) Reverse a Sub-list (medium) Reverse every K-element Sub-list (medium) 7. Pattern: Tree Breadth First Search,树上的BFS 这种模式基于宽搜(Breadth First Search (BFS)),适用于需要遍历一颗树。借助于队列数据结构,从而能保证树的节点按照他们的层数打印出来。打印完当前层所有...
LeetCode的数组 easy 题主要包括3题: 485,求最大的二进制元素1的连续值(最多个连续的1); 283; 27。 在Python种,数组最简单的可以通过 list 来实现,也可以在NumPy中使用。在LeetCode的算法题中,通常使用 list。 2 Python 中的数组实现和操作 list中的元素如果全部都是数字,那么就是一个数组。当然,Python中...
T1. 统计对称整数的数目(Easy) https://leetcode.cn/problems/count-symmetric-integers/ 题解(模拟) 根据题意模拟,亦可以使用前缀和预处理优化。 classSolution{funcountSymmetricIntegers(low:Int,high:Int):Int{varret=0for(xinlow..high){vals="$x"valn=s.lengthif(n%2!=0)continuevardiff=0for(iin0...
113-medium-C++-Path-Sum-II.md 114-medium-Flatten-Binary-Tree-to-Linked-List.md 120-medium-C++-Triangle 121-easy-C++-Best-time-to-buy-and-sell-stock 122-easy-C++-Best-Time-to-Buy-and-Sell-Stock-II 123-hard-C++-Best-Time-to-Buy-and-Sell-Stock-III.md ...
class Solution: def merge(self, nums1: List[int], m: int, nums2: List[int], n: int) -> None: """ Do not return anything, modify nums1 in-place instead. """ ans=[]fori inrange(len(nums1)):ifnums1[i]!=0:ans.append(i)forj inrange(len(nums2)):ifnums2[j]!=0:ans....
/*递归法class Solution {public:ListNode* reverseList(ListNode* head) {if (!head || !head->next)return head;ListNode* newHead = reverseList(head->next);head->next->next = head;head->next = nullptr;return newHead;}};*/// 迭代法classSolution{public:ListNode*reverseList(ListNode*head){Li...